417 + Interview Questions in PERMUTATIONS-AND-COMBINATIONS IN GENERAL AWARENESS Page 5 InterviewSolution

201.	A big contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag. If (i) they can be of any colour. (ii) two must be white and two red. (iii) they must all be of the same colour.
Answer» Total number of marbles = 6 white + 5 red = 11 marbles (i) If they can be of any colour means we have to select 4 marbles out of 11. `therefore` Required number of ways `=""^(11)C_(4)` (ii) If two must be white, then selection will be `""^(6)C_(2)` and two must be red, then selection will be `""^(5)C_(2)`. `therefore` Required number of ways `= ""^(6)C_(2) xx ""^(5)C_(2)` (iii) If they all must be of same colour, then selection of 4 white marbles out of `6 = ""^(6)C_(4)` and selection of 4 red marble out of `5 = ""^(5)C_(4)` `therefore` Required number of ways `=""^(6)C_(4) + ""^(5)C_(4)`

Discussion

202.	`""^(n)C_(n-r)+3.""^(n)C_(n-r+1)+3.""^(n)C_(n-r+2)+""^(n)C_(n-r+3)=""^(x)C_(r)`A. n+1B. n+2C. n+3D. n+4
Answer» Correct Answer - C

Discussion

203.	In how many ways can 5 red and 4 white balls be drawn from a bag containing 10 red and 8 white ballsA. `""^(8)C_(5)xx""^(10)C_(4)`B. `""^(10)C_(5)xx""^(8)C_(4)`C. `""^(18)C_(9)`D. none of these
Answer» Correct Answer - B

Discussion

204.	Find the sum of the series `(1^(2)+1)1!+(2^(2)+1)2!+(3^(2)+1)3!+ . .+(n^(2)+1)n!`.
Answer» Let `S_(n)=(1^(2)+1)1!+(2^(2)+1)2!+(3^(2)+1)3!+ . .+(n^(2)+1)n!` `therefore` nth term `T_(n)=(n^(2)+1)n!` `={(n+1)(n+2)-3(n+1)+2}n!` `T_(n)=(n+2)!-3(n+1)!+2n!` Putting `n=1,2,3,4, . .,n` then `T_(1)=3!-32!+21!` `T_(2)=4!-33!+22!` `T_(3)+%!-34!+23!` `T_(4)=6!-35!+24!` . . . . . . . . . . . . . . `T_(n-1)=(n+1)!-3n!+2(n-1)!` `T_(n)=(n+2)!-3(n+1)!+2n!` `therefore S_(n)=T_(1)+T_(2)+T_(3)+ . .. +T_(n)` `=(n+2)!-2(n+1)!` [the rest cancel out] `=(n+2)(n+1)!-2(n+1)!` `=(n+1)!(N=2-2)` `=n(n+1)!`

Discussion

205.	How many integers between 1 and 1000000 have the sum of the digit equal to 18?
Answer» Integers betweenn 1 and 1000000 will be 1,2,3,4,5 or 6 digits and given sum of digit=18 Thus, we need to obtain the number of solutions of the equation `x_(1)+x_(2)+x_(3)+x_(4)+x_(5)+x_(6)=18` . . . (i) where, `0 lt x_(i) le 9, i=1,2,3,4,5,6` Therefore, the number of solutions of Eq. (i) will be =Coefficient of `x^(18)` in `(x^(0)+x^(1)+x^(2)+x^(3)+ . . .+x^(9))^(6)` =Coefficient of `x^(18)` in `((1-x^(10))/(1-x))^(6)` =Coefficient of `x^(18)` in `(1-x^(10))^(6)(1-x)^(-6)` =Coefficient of `x^(18)` in `(1-6x^(10))(1+.^(6)C_(1)x+.^(7)C_(2)x^(2)+ . . .+.^(13)C_(8)x^(8)+ . . .+.^(23)C_(18)x^(18)+ . . .)` `=.^(23)C_(18)-6.^(13)C_(8)=.^(23)C_(5)-6.^(13)C_(5)` `=(2322212019)/(12345)-6(131211109)/(1234*5)` `=33649-7722=25927`

Discussion

206.	Find the negative terms of the sequence `X_(n)=(.^(n+4)P_(4))/(P_(n+2))-(143)/(4P_(n))`
Answer» We have, `x_(n)=(.^(n+4)P_(4))/(P_(n+2))-(143)/(4P_(n))` `thereforex_(n)=((n+4)(n+3)(n+2)(n+1))/((n+2)!)-(143)/(4n!)` `=((n+4)(n+3)(n+2)(n+1))/((n+2)(n+1)n!)-(143)/(4n!)` `=((n+4)(n+3))/(n!)-(143)/(4n!)=((4n^(2)+28n-95))/(4n!)` `because x_(n)` is negative `therefore((4n^(2)+28n-95))/(4n!) lt 0` which is true for n=1,2. Hence, `x_(1)=(63)/(4) and x_(2)=-(23)/(8)` are two negative terms.

Discussion

207.	What is the totalnumber of `2xx2`matrices with eachentry 0 or 1?A. 8B. 16C. 4D. none of these
Answer» Correct Answer - B

Discussion

208.	If all the letters of the word AGAIN be arranged as in a dictionary,what is the fiftieth word?A. NAAGIB. NAAIGC. NIAAGD. NAIAG
Answer» In dictionary the words at each stage are arranging the other four letters GAIN, we obtain 4!=24 words. Thus, there 24 words which start with A. These are the first 24 words. Then, starting with G, and arranging the other four letters A, A, I, N in different ways, we obtain `(4!)/(2!)=(24)/(2)=12` words. Thus, there are 12 words, which start with G. Now, we start with I. the remaining 4 letters A, G, A, N can be arranged in `(4!)/(4!)=12` ways. So, there are 12 words, which start with I, Thus, we have so far constructed 48 words. The 49th word is NAAGI and hence the 50th word is NAAIG.

Discussion

209.	How many numbers lying between `999` and `10000` can be formed with the help of the digit `0,2,3,6,7,8` when the digits are not to be repeatedA. 100B. 200C. 300D. 400
Answer» Correct Answer - C

Discussion

210.	How many four digit numbers can be formed using digits 1, 2, 3, 4, 5 such that at least one of the number is respeated?A. `4^(4)-5!`B. `4^(5)-4!`C. `5^(4)-4!`D. `5^(4)-5!`
Answer» Correct Answer - D

Discussion

211.	The total number of ways in which 12 persons can be divided into three groups of 4 persons each, isA. `(12!)/((13!)^(3)4!)`B. `(12!)/((4!)^(3))`C. `(12!)/((4!)^(3)3!)`D. `(12!)/((3!)^(4))`
Answer» Correct Answer - C

Discussion

212.	Number of 3 digit number in which digit at hundredth place is greater than other two ?
Answer» Total numbers of combinations `0^2+1^2+22^2+3^2+....+8^2` n=8 formula=`(n(2n+1)(n+1))/6` `=(8(8*2+1)(8+1))/6` `=194`.

Discussion

213.	In how many ways can five people be divided into three groups? (b) In how many ways can five people be distributed in three different rooms if no room must be empty? In how many ways can five people be arranged in three different rooms if no room must be empty and each room has 5 seats in a single row.
Answer» 5->1,1,3 `(5!)/(3!1!1!2!)=10` 5->1,2,2 `(5!)/(112!2!2!)=15` Total groups=25 c)5->1,1,3 =10 3 different rooms `5C_15C_15C_33!-(1)` 1,2,3=15 3 different rooms `5C_!5C_22!5C_22!-(2)` Total wages=adding 1 and 2 =90000+180000=270000.

Discussion

214.	Number of ways in which 12 different things can be distributed in 3 groups, isA. `(12!)/((4!)^(3))`B. `(12!)/(3!(4!)^(3))`C. `(12!)/(4!(3!)^(2))`D. `(12!)((3!)^(4))`
Answer» Correct Answer - B

Discussion

215.	The total number of ways of dividing `15` different things into groups of `8. ,4` and `3` respectively, isA. `(15!)/(8!4!(3!)^(2))`B. `(15!)/(8!4!3!)`C. `(15!)/(8!4!)`D. none of these
Answer» Correct Answer - B

Discussion

216.	Find the number of ways in which `n`different prizes can be distributed among `m(A. `n^(m)-n)`B. `m^(n)`C. mnD. none of these
Answer» Required of ways = Total number of ways - Number of ways in which one gets all the prizes `=m^(n)-m`

Discussion

217.	Let `n=180`. Find the number of positive integral divisors of `n^2`, which do not divide n
Answer» n=180=204 n=`2^23^25` `D_1=(2+1)(2+1)(1+1)-1` `D_1=17-(1)` `n^2=2^43^4*5^2` `D_2=(4+1)(4+1)(3)-2` `D_2=73-(2)` subtracting equation 1 from 2 =73-17=56 is our required number.

Discussion

218.	How many arrangements of the 9 letters a, b, c, p, q, r, x, y, z are there such that y is between x and z? (Any two, or all three, of the letters x, y, z may not be consecutive.)
Answer» Total possible ways`=.^7C_326!`

Discussion

219.	One hundred management students who read at least one of the three business magazines are surveyed to study the relationship pattern. its found that 80 read Business India, 50 read Business World and 30 read Business Today. Five students read all the three magazines. How many read exactly two magazines?
Answer» `=75-x-y+x+y+5` `=45-x-z+25-y-z` `=150-x-y-z=100` `=x+y+z=50`.

Discussion

220.	Let p be a prime number such that `p>=3`. Let `n=p!+1`. The number of primes in the list `n+1`, `n+2, n+3, ......n+p -1` is
Answer» When P=3,n=7 equation=(n+P+1)=(7+3-1)=9 When P=4,n=25 equation=(n+P+1)=(25+9-1)=28 When P=5,n=121 Perimeter=0 When P=6,n=721 Perimeter=0.

Discussion

221.	nP5=42(nP3), find n
Answer» `nC_55! =42nC_33!` `(n!)/(5!(n-5)!)5! =42(n!)/(3!(n-3)!3!` `((n-3)!)/((n-5)!)=42` `n^2-7n+12=42` `(n-10)(n+n)=0` `n=10`.

Discussion

222.	Find the number of 4 letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed.
Answer» required = `4321` `= 122 = 24` answer by fundamental principle of counting

Discussion

223.	There are 25 students in a class in which 15 boys and 10 girls. The class teacher select either a boy or girl for monitor of the class. In how many ways the class teacher can make this selection?
Answer» Since, there are 15 ways to select a boy, so there are 10 ways to select a girl. Hence, by the fundamental principal of addiition, either a boy or a girl can be performed in 15+10=25 ways.

Discussion

224.	Find n, if `(n+2)=60xx(n-1)!`
Answer» `because(n+2)!=(n+2)(n+1)n(n-1)!` `implies((n+2)!)/((n-1)!)=(n+2)(n+1)n` `implies60=(n+2)(n+1)n` [given] `implies5xx4xx3=(n+2)xx(n+1)xxn` `thereforen=3`

Discussion

225.	Evaluate `sum_(r=1)^(n)rxxr!`
Answer» We have, `underset(r=1)overset(n)(sum)rxxr!=underset(r=1)overset(n)(sum){(r+1)-1}r!=underset(r=1)overset(n)(sum)(r+1)!-r!` `=(n+1)!-1!` [put r=n in (r+1)! Annd r=1 is r!] `=(n+1)!-1`

Discussion

226.	In how many ways five different rings can be put on four fingers with at least one ring on each finger?A. 120B. 480C. 240D. 960
Answer» Since at least one ring is to be put in each finger. So, we select four rings out of 5 and then they are arranged in fingers. This can be done in `""^(5)C_(4)xx4!` ways. Now, either above the ring already placed on it or below it. So, the required number of ways `=""^(5)C_(4)xx4!xx4xx2=960`

Discussion

227.	In how many ways five different rings can be worn in four fingers with at least one ring in each finger?A. 120B. 96C. 20D. 480
Answer» Since at least one ring is to be worn in each finger. So, we first select four rings out of 5 given rings and then they are arranged in fingers. This can be done in `""^(5)C_(4)xx4!` ways. Now, one ring is left which can be worn in any one of the four fingers in 4 ways. Hence, required number of ways `=""^(5)C_(4)xx4!xx4=480`.

Discussion

228.	The total number not more than 20 digits that areformed by usingd the digits 0, 1, 2, 3, and 4 isa. `5^(20)`b. `5^(20)-1`c. `5^(20)+1`d.none of theseA. `5^(20)`B. `5^(20)-1`C. `5^(20)+1`D. none of these
Answer» We have, Number of numbers `=sum_(r=1)^(20)` (No. of r digit numbers) `=5+4.5+4.5^(2)+4.5^(3)+......+4.5^(19` `=5+4xx5((5^(19)-1)/(5-1))=5+5^(30)-5=5^(20)` ALITER Number of numbers `=sum_(r=1)^(20)` (N0. of r digit numbers) `=5+sum_(r=1)^(20)4xx5^(r-1)`

Discussion

229.	Which of the following is incorrect?A. `""^(n)C_(r)=""^(n)C_(n-r)`B. `""^(n)C_(r)=""^(n-1)C_(r)+""^(n)C_(n-r)`C. `""^(n)C_(r)=""^(n-1)C_(r)+""^(n-1)C_(n-r)`D. `r!""^(n)C_(r)=""^(n)P_(n)`
Answer» Correct Answer - B

Discussion

230.	By using the digits 0,1,2,3,4 and 5 (repetitions not allowed) numbersare formed by using any number of digits. Find the total number of non-zeronumber that can be formed.A. 1030B. 1630C. 1200D. 1530
Answer» Number of numbers= Number of 1 digit number + No. of 2 digit numbers +……+ Number of 6 digit numbers `=5+5xx5+5xx4xx5xx5xx4xx3+5xx5xx4xx3xx2+5xx5xx4xx3xx2xx1` `=5+25+100+300+600+600=1630`

Discussion

231.	ABCD is a convex quadrilateral and 3, 4, 5, and 6points are marked on the sides AB, BC, CD, and DA, respectively. The numberof triangles with vertices on different sides isa. `270`b. `220`c. `282`d. `342`A. 220B. 270C. 282D. 342
Answer» Correct Answer - D Total number of triangles=Number of triangles with vertices on sides `[(AB,BC,CD)+(AB,BC,DA)+(AB,CD,DA)+(BC,CD,DA)]` `=.^(3)C_(1)xx.^(4)C_(1)xx.^(5)C_(1)+.^(3)C_(1)xx.^(4)C_(1)xx.^(6)C_(1)+.^(3)C_(1)xx.^(5)C_(1)xx.^(6)C_(1)+.^(4)C_(1)xx.^(5)C_(1)xx.^(6)C_(1)` `=60+72+90+120=342`

Discussion

232.	How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that(i) repetition of the digits is allowed?(ii) repetition of the digits is not allowed?
Answer» (i) `---` `= 555= 125` (ii) `---` `543 = 60` answer

Discussion

233.	There are 4 mangoes, 3 apples, 2 oranges and 1 each of 3 other varieties of fruits. The number of ways of selecting at least one fruit of each kind, isA. 10!B. 9!C. 4!D. none of these
Answer» We, have, Required number of ways `=4xx3xx2xx1xx1xx1=4!`

Discussion

234.	There are 4 oranges, 5 apples and 6 mangoes in a fruit basket. In how manyways can a person make a selection of fruits from among the fruits in the basket?
Answer» Zero or more oranges can be selected out of 4 identical oranges=4+1=5 ways. Similarly, for apples number of selection=5+1=6 ways and mangoes can be selected in 6+1=7 ways. `therefore`The total number of selection when all the three kinds of fruits are selected`=5xx6xx7=210` But, in one of these selection number of each kind of fruit is zero and hence this selectionn must be excluded. `therefore`Required number=210-1=209

Discussion

235.	Consider the set of eight vector `V={a hat i+b hat j+c hat k ; a ,bc in {-1,1}}dot`Threenon-coplanar vectors can be chosen from `V`is `2^p`ways. Then `p`is_______.
Answer» Correct Answer - 5 Given 8 vectors are (1,1,1),(-1,1,1),(1,-1,1),(1,1,-1),(-1,-1,1),(1,-1,-1),(-1,1,-1),(-1,-1,-1) there are 4 diagonals of a cube. Now, for 3 non-coplanar vectors first we select 3 groups of diagonals and its opposite in `.^(4)C_(3)=4` ways. Then one vector from each group cann be selected in `2xx2xx2=8` ways. `therefore`Total ways`=4xx8=32=2^(5)=2^(p)` (given) Hence, `p=5`

Discussion

236.	Let `ngeq2`be integer. Take `n`distinct points on a circle and join each pair of pointsby a line segment. Color the linesegment joining every pair of adjacent points by blue and the rest by red. Ifthe number of red and blue line segments are equal, then the value of `n`is
Answer» Number of adjacent lines=n Number of non-adjacent lines`=.^(n)C_(2)-n=(n(n-3))/(2)` `therefore (n(n-3))/(2)=n implies(n(n-5))/(2)=0 impliesn=0` or 5 but `2 ge 2 impliesn=5`

Discussion

237.	Let `n_1
Answer» Correct Answer - 7 If `n_(1),n_(2),n_(3),n_(4)` take minimum values 1,2,3,4 respectively, then `n_(5)` will be maximum 10. `therefore`corresponding to `n_(5)=10` there is only one solution `n_(1)=1,n_(2)=2,n_(3)=3,n_(4)=4` Corresponnding to `n_(5)=9`, we can have, `n_(1)=1,n_(2)=2,n_(3)=3,n_(4)=5` i.e., one solution Corresponding to `n_(5)=8`, we have, `n_(1)=1,n_(2)=2,n_(3)=3,n_(4)=6` or `n_(1)=1,n_(2)=2,n_(3)=4,n_(4)=5` i.e., two solutions corresponding to `n_(5)=5`, we can have `n_(1)=1,n_(2)=2,n_(3)=4, n_(4)=6` or `n_(1)=1,n_(2)=3,n_(3)=4,n_(4)=5` i.e., two solutions Corresponding to `n_(5)=6` we can have `n_(1)=2,n_(2)=3,n_(3)=4,n_(4)=5` i.e., one solution thus, there can be 7 solutions.

Discussion

238.	Let A and B be two sets containing 2 elements and 4elements respectively. The number of subsets of `AxxB`having 3 ormore elements is(1) 220 (2)219(3) 211 (4)256
Answer» total elements =`2 xx4 = 8` total subsets= `2^8` subset having 0 elements= `.^8c_0` subset having 1 element=`.^8c_1` subset having 2 element`= .^8c_2` `n(A XXB)= 2^8 - .^8c_0 - .^8c_1 - .^8c_2` `= 2^8- [1 + 8 + (8 xx7)/2]` `= 2^8 -[9+28]` `= 2^8- 37` `= 256-37= 219` option 2 is correct

Discussion

239.	Let `S={1,2,3ddot,9}dotFork=1,2, 5,l e tN_k`be the number of subsets of S, each containing five elements out ofwhich exactly `k`are odd. Then `N_1+N_2+N_3+N_4+N_5=?`210 (b)252 (c) 125(d) 126A. 210B. 252C. 126D. 125
Answer» Correct Answer - C `N_(i) = ""^(5)C_(k) xx ""^(4)C_(5 - k)` `N_(1) = 5 xx 1` `N_(2) = 10 xx 4` `N_(3) = 10 xx 6` `N_(4) = 5 xx 4` `N_(5) = 1` `N_(1) + N_(2) + N_(3) + N_(4) + N_(5) = 126`

Discussion

240.	Let `T_n`be the numberof all possible triangles formed by joining vertices of an n-sided regularpolygon. If `T_(n+1)-T_n=""10`, then thevalue of n is(1) 5(2) 10(3) 8(4) 7A. 7B. 5C. 10D. 8
Answer» Correct Answer - B Given, `T_(n) = ""^(n)C_(3) rArr T_(n + 1) = ""^(n + 1)C_(3)` `therefore" " T_(n + 1) - T_(n) = ""^(n + 1)C_(3) = ""^(n)C_(3) = 10" "` [given] `rArr ""^(n)C_(2) + ""^(n)C_(3) - ""^(n)C_(3) = 10 " "[because ""^(n)C_(r) + ""^(n)C_(r + 1) = ""^(n + 1)C_(r + 1)]` `rArr " "^(n)C_(2) = 10` `rArr` n = 5

Discussion

241.	A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal toA. 28B. 27C. 25D. 24
Answer» Correct Answer - C It is given that the group of students comprises of 5 boys and n girls. The number of ways, in which a team of 3 students can be selected from this group such that each team consists of at least one boy and at least one girls, is = (number of ways selecting one boy and 2 girls) + (number of ways selecting two boys and 1 girl) `= (""^(5)C_(1) xx ""^(n)C_(2)) (""^(5)C_(2) xx ""^(n)C_(1)) = 1750` [given] `rArr " " (5 xx (n(n - 1))/(2)) + ((5 xx 4)/(2) XX n) = 1750` `rArr n(n - 1) + 4n = 2/5 xx 1750 rArr n^(2) + 3n = 2 xx 350` `rArr n^(2) + 3n - 700 = 0`

Discussion

242.	A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team isA. 380B. 320C. 260D. 95
Answer» Correct Answer - A Either one boy will be selected or no boy will be selected. Also out of four members one captain is to be selected. `therefore`Required number of ways`=(.^(4)C_(1)xx.^(6)C_(3)+.^(6)C_(4))xx.^(4)C_(1)` `=(4xx20+15)xx4=95xx4=380`

Discussion

243.	The number of ways in which 10 condidates `A_(1),A_(2),......,A_(10)` can be ranked so that `A_(1)` is always above `A_(2)`, isA. 10!B. `(10!)/(2)`C. 9!D. none of these
Answer» Ten candidates can be ranked in 10! ways. In half of these ways A1 is above `A_(2)` and in another half `A_(2)` is above `A_(1)`. So, required number of ways `=(10!)/(2)`

Discussion

244.	The number of ways in which a team of 11 players can be selected from 22 players .including 2 of them and excluding 4 of them isA. `""^(16)C_(11)`B. `""^(16)C_(5)`C. `""^(16)C_(9)`D. `""^(20)C_(9)`
Answer» Correct Answer - C

Discussion

245.	The number of ways in which a team of 11 players can be selected from 22 players including 2 of them and excluding 4 of them isA. `""^(16)C_(11)`B. `""^(16)C_(5)`C. `""^(16)C_(9)`D. `""^(20)C_(9)`
Answer» Correct Answer - C Total number of players = 22 We have to select a team of 11 players. Selection of 11 players when 2 of them is always included and 4 are never included. Total number of players = 22 - 2 - 4 = 16 `therefore` Required number of selection `= ""^(16)C_(9)`

Discussion

246.	If 7 points out of 12 are in the same straight line, then the number of triangles formed isA. 19B. 158C. 185D. 201
Answer» Correct Answer - C

Discussion

247.	In an examination, there are three multiple choice questions and each question has four choices. Number of ways in which a student can fail to get all answers correct, isA. 11B. 12C. 27D. 63
Answer» Correct Answer - D

Discussion

248.	In how many ways can a team 11 players be formedout of 25 players, if 6 out of them are always to be included and 5 always tobe excludeda. `2020`b. `2002`c. `2008`d. `8002`A. 2002B. 2008C. 2020D. 8002
Answer» Correct Answer - A

Discussion

249.	The number of ways of dividing 20 persons into 10 couples isA. `(20!)/(2^(10))`B. `""^(20)C_(10)`C. `(20!)/((21!)^(9))`D. none of these
Answer» Here, the order of the couples is not important. So, required number of ways `=(20!)/(2^(10)10!)`

Discussion

250.	In how many ways, we can choose two teams of mixed double for a tennis tournament from four couples such that if any couple participates, then it is in the same team?
Answer» Correct Answer - 42 Case I when no couple is chosen We can choosen in `.^(4)C_(2)=6` ways and hence two teams can be formed in `2xx6=12` ways. Case II when only one couple is chosen A couple can be chosen in `.^(4)C_(1)=4` ways and the other team can be chosen in `.^(3)C_(1)xx.^(2)C_(1)=6` ways. hence, two teams can be formed in `4xx6=24` ways. Case III when two couples are chosen. then team can be chosen in `.^(4)C_(2)=6` ways. hence, total ways=12+24+6=42.

Discussion

Explore topic-wise InterviewSolutions in .

A big contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag. If (i) they can be of any colour. (ii) two must be white and two red. (iii) they must all be of the same colour.

`""^(n)C_(n-r)+3.""^(n)C_(n-r+1)+3.""^(n)C_(n-r+2)+""^(n)C_(n-r+3)=""^(x)C_(r)`A. n+1B. n+2C. n+3D. n+4

In how many ways can 5 red and 4 white balls be drawn from a bag containing 10 red and 8 white ballsA. `""^(8)C_(5)xx""^(10)C_(4)`B. `""^(10)C_(5)xx""^(8)C_(4)`C. `""^(18)C_(9)`D. none of these

Find the sum of the series `(1^(2)+1)1!+(2^(2)+1)2!+(3^(2)+1)3!+ . .+(n^(2)+1)n!`.

How many integers between 1 and 1000000 have the sum of the digit equal to 18?

Find the negative terms of the sequence `X_(n)=(.^(n+4)P_(4))/(P_(n+2))-(143)/(4P_(n))`

What is the totalnumber of `2xx2`matrices with eachentry 0 or 1?A. 8B. 16C. 4D. none of these

If all the letters of the word AGAIN be arranged as in a dictionary,what is the fiftieth word?A. NAAGIB. NAAIGC. NIAAGD. NAIAG

How many numbers lying between `999` and `10000` can be formed with the help of the digit `0,2,3,6,7,8` when the digits are not to be repeatedA. 100B. 200C. 300D. 400

How many four digit numbers can be formed using digits 1, 2, 3, 4, 5 such that at least one of the number is respeated?A. `4^(4)-5!`B. `4^(5)-4!`C. `5^(4)-4!`D. `5^(4)-5!`

The total number of ways in which 12 persons can be divided into three groups of 4 persons each, isA. `(12!)/((13!)^(3)4!)`B. `(12!)/((4!)^(3))`C. `(12!)/((4!)^(3)3!)`D. `(12!)/((3!)^(4))`

Number of 3 digit number in which digit at hundredth place is greater than other two ?

Number of ways in which 12 different things can be distributed in 3 groups, isA. `(12!)/((4!)^(3))`B. `(12!)/(3!(4!)^(3))`C. `(12!)/(4!(3!)^(2))`D. `(12!)((3!)^(4))`

The total number of ways of dividing `15` different things into groups of `8. ,4` and `3` respectively, isA. `(15!)/(8!4!(3!)^(2))`B. `(15!)/(8!4!3!)`C. `(15!)/(8!4!)`D. none of these

Find the number of ways in which `n`different prizes can be distributed among `m(A. `n^(m)-n)`B. `m^(n)`C. mnD. none of these

Let `n=180`. Find the number of positive integral divisors of `n^2`, which do not divide n

How many arrangements of the 9 letters a, b, c, p, q, r, x, y, z are there such that y is between x and z? (Any two, or all three, of the letters x, y, z may not be consecutive.)

Let p be a prime number such that `p>=3`. Let `n=p!+1`. The number of primes in the list `n+1`, `n+2, n+3, ......n+p -1` is

nP5=42(nP3), find n

Find the number of 4 letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed.

There are 25 students in a class in which 15 boys and 10 girls. The class teacher select either a boy or girl for monitor of the class. In how many ways the class teacher can make this selection?

Find n, if `(n+2)=60xx(n-1)!`

Evaluate `sum_(r=1)^(n)rxxr!`

In how many ways five different rings can be put on four fingers with at least one ring on each finger?A. 120B. 480C. 240D. 960

In how many ways five different rings can be worn in four fingers with at least one ring in each finger?A. 120B. 96C. 20D. 480

The total number not more than 20 digits that areformed by usingd the digits 0, 1, 2, 3, and 4 isa. `5^(20)`b. `5^(20)-1`c. `5^(20)+1`d.none of theseA. `5^(20)`B. `5^(20)-1`C. `5^(20)+1`D. none of these

Which of the following is incorrect?A. `""^(n)C_(r)=""^(n)C_(n-r)`B. `""^(n)C_(r)=""^(n-1)C_(r)+""^(n)C_(n-r)`C. `""^(n)C_(r)=""^(n-1)C_(r)+""^(n-1)C_(n-r)`D. `r!""^(n)C_(r)=""^(n)P_(n)`

By using the digits 0,1,2,3,4 and 5 (repetitions not allowed) numbersare formed by using any number of digits. Find the total number of non-zeronumber that can be formed.A. 1030B. 1630C. 1200D. 1530

ABCD is a convex quadrilateral and 3, 4, 5, and 6points are marked on the sides AB, BC, CD, and DA, respectively. The numberof triangles with vertices on different sides isa. `270`b. `220`c. `282`d. `342`A. 220B. 270C. 282D. 342

How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that(i) repetition of the digits is allowed?(ii) repetition of the digits is not allowed?

There are 4 mangoes, 3 apples, 2 oranges and 1 each of 3 other varieties of fruits. The number of ways of selecting at least one fruit of each kind, isA. 10!B. 9!C. 4!D. none of these

There are 4 oranges, 5 apples and 6 mangoes in a fruit basket. In how manyways can a person make a selection of fruits from among the fruits in the basket?

Consider the set of eight vector `V={a hat i+b hat j+c hat k ; a ,bc in {-1,1}}dot`Threenon-coplanar vectors can be chosen from `V`is `2^p`ways. Then `p`is_______.

Let `ngeq2`be integer. Take `n`distinct points on a circle and join each pair of pointsby a line segment. Color the linesegment joining every pair of adjacent points by blue and the rest by red. Ifthe number of red and blue line segments are equal, then the value of `n`is

Let `n_1

Let A and B be two sets containing 2 elements and 4elements respectively. The number of subsets of `AxxB`having 3 ormore elements is(1) 220 (2)219(3) 211 (4)256

Let `S={1,2,3ddot,9}dotFork=1,2, 5,l e tN_k`be the number of subsets of S, each containing five elements out ofwhich exactly `k`are odd. Then `N_1+N_2+N_3+N_4+N_5=?`210 (b)252 (c) 125(d) 126A. 210B. 252C. 126D. 125

Let `T_n`be the numberof all possible triangles formed by joining vertices of an n-sided regularpolygon. If `T_(n+1)-T_n=""10`, then thevalue of n is(1) 5(2) 10(3) 8(4) 7A. 7B. 5C. 10D. 8

A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal toA. 28B. 27C. 25D. 24

A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team isA. 380B. 320C. 260D. 95

The number of ways in which 10 condidates `A_(1),A_(2),......,A_(10)` can be ranked so that `A_(1)` is always above `A_(2)`, isA. 10!B. `(10!)/(2)`C. 9!D. none of these

The number of ways in which a team of 11 players can be selected from 22 players .including 2 of them and excluding 4 of them isA. `""^(16)C_(11)`B. `""^(16)C_(5)`C. `""^(16)C_(9)`D. `""^(20)C_(9)`

The number of ways in which a team of 11 players can be selected from 22 players including 2 of them and excluding 4 of them isA. `""^(16)C_(11)`B. `""^(16)C_(5)`C. `""^(16)C_(9)`D. `""^(20)C_(9)`

If 7 points out of 12 are in the same straight line, then the number of triangles formed isA. 19B. 158C. 185D. 201

In an examination, there are three multiple choice questions and each question has four choices. Number of ways in which a student can fail to get all answers correct, isA. 11B. 12C. 27D. 63

In how many ways can a team 11 players be formedout of 25 players, if 6 out of them are always to be included and 5 always tobe excludeda. `2020`b. `2002`c. `2008`d. `8002`A. 2002B. 2008C. 2020D. 8002

The number of ways of dividing 20 persons into 10 couples isA. `(20!)/(2^(10))`B. `""^(20)C_(10)`C. `(20!)/((21!)^(9))`D. none of these

In how many ways, we can choose two teams of mixed double for a tennis tournament from four couples such that if any couple participates, then it is in the same team?