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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
How many automobile license plates can be made, if each plate contains two different letters followed by three different digits ? |
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Answer» Correct Answer - 15552 There are 26 English alphabets and 10 digits (0 to 9). Since, it is given that each plate contains two different letters followed by three different digits. `therefore` Arrangement of 26 letters, taken 2 at a time `= .^(26)P_(2)` `=26xx25` =650 And arrangement of 10 digit taken three at a time `= .^(10)P_(3)` `=10xx9xx8` =720 `therefore` Total number of license plates `=650xx720=468000` |
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| 152. |
Find the number of ways of dividing 6 couples in 3 groups if each group has exactly one couple and each group has 2 males and 2 females. |
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Answer» There are 6 couples `(A_(1)B_(1)),(A_(2)B_(2)),(A_(3)B_(3)),(A_(4)B_(4)),(A_(5)B_(5)) " and " (A_(6)B_(6)), " where " A_(i)` is male and `B_(i)` is female. We have to divide these couples (12 persons) in three groups P, Q and R, such that each containing exactly one couple having two males and two females. First select three couples for three groups P, Q, R. Number of ways are `""^(6)C_(3)`. Now from the remaining three couples three males can enter in groups P, Q and R in 3! ways. The remaining three females cannot enter the group in which their spouses have entered. So, for females we have derangement `D(3)=3!(1-(1)/(1!)+(1)/(2!)-(1)/(3!))=2` Therefore, total number of required ways `= ""^(6)C_(3)xx3!xx2` =240 |
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| 153. |
In how many ways can the letters of the word PERMUTATIONS bearranged if the(i) words start with P and end with S,(ii) vowels are all together,(iii) there are always 4 letters between P andS? |
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Answer» (i) There are remaining 10 letters in the words starting with P and ending with S in which T occurs twice. No. of such words `= (10!)/(2!) = 1814400`. (ii) If all vowels A,E,I,O,U are together then taking them as on letter, the number of ways of arranging, 8 letters in which T occurs twice `=(8!)/(2!) = (40320)/(2) = 20160` Again, no. of ways of arranging 5 vowels `.^(5)P_(5) = 5! =120` Therefore, total words in which vowels are together `= 20160 xx 120 = 2419200` (iii) The 12 letters of the word PERMUTATIONS can filled at 12 places in the following way: 1 2 3 4 5 6 7 8 9 10 11 12 If there are always four letters between P and S then we can place P at 1,2,3,4,5,6, or 7th place and as a result S can be placed at 6,7,8,9,10,11 or 12th place. So, no of ways to place `P = 7` No. of ways to place `S = 7` Total ways `= 7 +7 14` No. of ways to place remaining 10 letterss at remaining 10 places in which T occur twice. `= (10!)/(2!) = (3628800)/(2) = 1814400` Therefore, required number of words `= 1814400 xx 14` `= 25401600` |
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| 154. |
Find all 3 digit numbers formed with the digits 1,2,3,4 and 5 if (i) repetition of digits is allowed? (ii) repetition of digits is not allowed? |
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Answer» (i) When repetition of digits is allowed: No. of ways of selecting 1st digit = 5 No. of ways of selecting 2nd digit = 5 No. of ways of selecting 3rd digit = 5 `:.` Total no of ways `= 5 xx 5 xx 5 = 125` (ii) When repetition of digits is not allowed. No. of ways of selecting 1st digit = 5 No. of ways of selecting 2nd digit = 4 No. of ways of selecting 3rd digit = 3 `:.` Total number of ways `= 5 xx 4 xx 3 = 60` |
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| 155. |
Determine thenumber of 5 card combinations out of a deck of 52 cards if there is exactlyone ace in each combination. |
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Answer» Since, there is exactly one ace in a combination of 5 cards, so no of ways of selecting one ace `= .^(4)C_(1) = 4` Again, no. of ways of selecting remaining 4 cards from remaining 48 cards `= .^(48)C_(4) = (48 xx 47 xx 46 xx 45)/(4 xx 3 xx 2xx 1) = 194580` Therefore, number of total combinations `= 194580 xx 4 = 778320` |
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| 156. |
Out of 10 white, 9 black, and 7 red balls, find the number of ways inwhich selection of one or more balls can be made (balls of the same color areidentical). |
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Answer» Correct Answer - 879 The required number of ways is : (10+1)(9+1)(7+1)-1 `=11xx10xx8-1=879` |
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| 157. |
In how many ways can 7 green and 5 yellowballs be arranged in a striaght line when 2 yellow balls are not together? |
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Answer» Two yellow balls never occur together, so first we will arrange green balls in a row. All green balls are similar, so they can arrange in a row in one way. `xx G xx G xx G xx G xx G xx G xx G xx` Now we have `7 +1 = 8` places (x) at which we can arrange the yellow balls. In this case, no two yellow balls be together. `:.` No. of ways to arrange 5 yellow balls at 5 places out of 8 places `= .^(8)C_(5) = 56` and required number of ways `= 1 xx 56 = 56`. |
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| 158. |
Determine thenumber of 5-card combinations out of a deck of 52 cards if each selection of5 cards has exactly one king. |
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Answer» No. of ways of selecting 1 king out of 4 kings `= .^(4)C_(1) = 4` No. of ways of selecting 4 cards from the remaining 48 cards `= .^(48)C_(4)` `= (48 xx 47 xx 46 xx 45)/(4 xx 3 xx 2 xx 1) = 194580` So, number of total groups `= 4 xx 194580 = 778320`. |
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| 159. |
How many signals can be formed with 5 different colours flags, if there are two flags in every signal? |
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Answer» No. of results for first flag = 5 No. of results for second flag = 4 `:.` Total required signals `= 5 xx 4 = 20` |
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| 160. |
Three are 6 different flags. Find the number of signals which can be formed with the help of at least 3 flags. |
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Answer» No. of flag required for one signal =3 or 4 or 5 or 6. Now number of ways to fill 3 places from 6 flags `= 6 xx 5 xx 4 = 120` Similarly, No. of signals containing flags `= 6 xx 5 xx 4 xx 3 = 360` No. of signals containing 5 flags `= 6 xx 5 xx 4 xx 3 xx 2 = 720` No. of signals containing 6 flags `= 6 xx 5 xx 4 xx 3 xx 2 xx 1 = 720` Total flags `= 120 _ 360 +720 +720 = 1920`. |
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| 161. |
The English alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed from the alphabet? |
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Answer» No. of ways of selecting 2 vowels out of 5 vowels `= .^(5)C_(2)`. No. of ways of selecting 2 consonants out of 21 consonants `= .^(21)C_(2)`. No. of ways to arrange 2 vowels and 2 consonants i.e., 4 letters = 4! Therefore total no. of words `= .^(5)C_(2) xx .^(21)C_(2) xx 4!` `= (5 xx 4)/(2 xx 1) xx (21 xx 20)/(2 xx 1) xx 24` `= 10 xx 210 xx 24` `= 50400` |
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| 162. |
Find number of negative integral solution of equation `x + y + z = - 12`A. `44`B. `55`C. `66`D. none of these |
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Answer» Correct Answer - B `(b)` `x+y+z=-12` Here `-10 le x`, `y`, `z le -1` Let `x_(1)=-x`, `y_(1)=-y`, `z_(1)=-z` `:.` we have `x_(1)+y_(1)+z_(1)=12` , where `1 le x_(1)`, `y_(1)`, `z_(1) le 12` `:.` Number of solutions of the equation `=^(12-1)/(C_(3-1)=^(11)C_(2)=55` |
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| 163. |
Find the exponent of 80 in 200!. |
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Answer» Correct Answer - 49 `80=2^(4)xx5` To find the exponent of 80 in 200!, we find the exponent of 2 and 5. Exponent of 2 is `[(200)/(2)]+[(200)/(2^(2))]+[(200)/(2^(3))]+[(200)/(2^(4))]+[(200)/(2^(5))]+[(200)/(2^(6))]+[(200)/(2^(7))]` `=100+50+25+12+6+3+1` =197 Exponent of 5 of `[(200)/(5)]+[(200)/(5^(2))]+[(200)/(5^(3))]=40+8+1=49` Now, exponent of 16 in 200! is `[197//4]=49`. Hence, exponent of 80 is 49. |
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| 164. |
The number of ways in which `p+q` things can be divided into two groups containing `p and q` things respectively isA. `((p+q+r)!)/(p! q! r!)`B. `((pqr)!)/((p+q+r)!)`C. `((p+q+r)!)/((pqr)!)`D. None of these |
| Answer» Correct Answer - A | |
| 165. |
The number of ways can five people be divided into three groups isA. `20`B. `25`C. `30`D. `36` |
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Answer» Correct Answer - B `(b)` Five people can be divided into three groups in `(1,1,3)` or `(1,2,2)` ways Hence, total number of ways `=(5!)/(3!(1!)^(2))xx(1)/(2!)+(5!)/((2!)^(2)1!)xx(1)/(2!)` `=10+15=25` |
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| 166. |
Find the remainder when `1!+2!+3!+4!++n !`is divided by 15, if `ngeq5.` |
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Answer» Correct Answer - 3 Let, `N=1!+2!+3!+4!+5!+6!+..+n!` `implies (N)/(15)=(1!+2!+3!+4!+5!+..+n!)/(15)` `=(1!+2!+3!+4!)/(15)+(5!+6!+..+n!)/(15)` `=(33)/(15)`+integer (as 5!,6!,.. Are divisible by 15) Hence , remainder is 3. |
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| 167. |
Prove that `(n !+1)`is not divisible by any natural number between `2a n dndot` |
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Answer» Let p be divisible by k and r be any natural number between 1 and k. If p+r is divided by k, we obtian r as the remainder. Now, `n!=1xx2xx3xx4xx..xx(n-1)n` Therefore, n! is divisible by every natural number between 2 and n. So, n!+1, when divided by any natural number between 2 and n, leaves 1 as the remainder. Hence, n!+1 is not divisible by any natural number between 2 and n. |
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| 168. |
Find the number of ways of dividing 52 cards among four playersequally. |
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Answer» First divide 52 cards equally into four groups, the number of ways is `(52!)/((13!)^(4)(4!))` Now this four groups can be distributed among four players in 4! Ways. Then total number of ways of dividing among four players equally is `(52!)/((13!)^(4)(4!))xx4!=(52!)/((13!)^(4))`. |
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| 169. |
In how many ways can Rs. 16 be divided into 4 persons when none of themgets less than Rs. 3? |
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Answer» Correct Answer - 35 First give Rs 3 to each of four person. Now for remaining Rs 4 we have x+y+z+w=4 where x, y, z, w are number of rupees gained by person 1,2,3,4, respectively. We have to find number of non-negative solutions of Eq. (1), which is `. (4+4-1)C_(4-1)= .^(7)C_(3)=35`. |
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| 170. |
What is `((n + 2)! + (n + 1) (n -1)!)/((n +1) (n -1)!)` equal to ?A. 1B. Always an odd integerC. A perfect squareD. None of the above |
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Answer» Correct Answer - C Given expression is : `((n +2) ! + (n +1)!(n -1)!)/((n +1)! (n -1)!) = x` (let) `rArr x = ((n +2) (n +1) n (n -1) ! + (n +1) (n -1)!)/((n +1) (n -1)!)` `= (n +2) n + 1 = n^(2) + 2n + 1 = (n +1)^(2)` Which is a perfect square |
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| 171. |
A meeting is to be addressed by 5 speakers A, B, C, D, E. In how many ways can the speakers be ordered, if B must not precede A (immediately or otherwise)?A. 120B. 24C. 60D. `5^(4) xx 4` |
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Answer» Correct Answer - B According to given restriction. B must not preced A (immediately or otherwise) `rArr` A must follow B, i.e., B should addressed the meeting at first place. So, rest of the four speakers can address in 4! Ways. `:.` Required number of ways `= 4! = 24` |
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| 172. |
Find the number of ways of selecting 10 balls out o fan unlimitednumber of identical white, red, and blue balls. |
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Answer» Correct Answer - `.^(12)C_(2)` Let `x_(W),x_(R ),x_(B)` be the number of white balls, red balls, and blue balls being selected. We must have `x_(W)+x_(R )+x_(B)=10` The required number of ways is equal to the number of non-negative integers solutions of `x_(W)+x_(R )+x_(B)=10`, which is `.(3+10-1)C_(10)= .^(12)C_(10)= .^(12)C_(2)`. |
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| 173. |
How many ordered pairs of (m,n) integers satisfy `(m)/(12)=(12)/(n)`?A. `30`B. `15`C. `12`D. `10` |
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Answer» Correct Answer - A `(a)` `(m)/(12)=(12)/(n)` `impliesmn=144impliesn=(144)/(m)` We want to find here total number of divisors of `144` `144=2^(4)3^(2)` Total divisors are `(4+1)(2+1)=15` But negative pairs are also possible hence `=30` |
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| 174. |
If x, y, z, t are odd natural numbers such that `x + y + z +w=20` then find the number of values of ordered quadruplet (x, y, z, t). |
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Answer» Correct Answer - 165 Let a=2p+1, b=2p+1,c=2r+1,d=2s+1 where p,q,r,s are non-negative integers. Therefore, 2p+1+2p+1+2r+1+2s+1=20 or p+q+r+s=8 Therefore, the required number is equal to the number of non-negative intergral solutions of p+q+r+s=8, which is given by `.(8+4-1)C_+(4-1)= .^(11)C_(3)=165`. |
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| 175. |
Find the maximum number of points of intersection of 6 circles. |
| Answer» Two circles intersect maximum at two distinct points. Now, two circles can be selected in `.^(6)C_(2)` ways. Again, each selection of two circles gives two points of intersection. Therefore, the total number of points of intersection is `.^(6)C_(2)xx2=30`. | |
| 176. |
Let `theta=(a_(1),a_(2),a_(3),...,a_(n))` be a given arrangement of `n` distinct objects `a_(1),a_(2),a_(3),…,a_(n)`. A derangement of `theta` is an arrangment of these `n` objects in which none of the objects occupies its original position. Let `D_(n)` be the number of derangements of the permutations `theta`. The relation between `D_(n)` and `D_(n-1)` is given byA. `D_(n)-nD_(n-1)=(-1)^(n)`B. `D_(n)-(n-1)D_(n-1)=(-1)^(n-1)`C. `D_(n)-nD_(n-1)=(-1)^(n-1)`D. `D_(n)-D_(n-1)=(-1)^(n-1)` |
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Answer» Correct Answer - A `(a)` `D_(n)-nD_(n-1)=(-1)(D_(n-1)-(n-1)D_(n-2))` By implied induction on `n`, we obtain `D_(n)-nD_(n-1)=(-1)^(n-2)(D_(2)-2D_(1))`, Where `D_(1)=0` and `D_(2)=1` `=(-1)^(n)` |
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| 177. |
Let `theta=(a_(1),a_(2),a_(3),...,a_(n))` be a given arrangement of `n` distinct objects `a_(1),a_(2),a_(3),…,a_(n)`. A derangement of `theta` is an arrangment of these `n` objects in which none of the objects occupies its original position. Let `D_(n)` be the number of derangements of the permutations `theta`. There are `5` different colour balls and `5` boxes of colours same as those of the balls. The number of ways in which one can place the balls into the boxes, one each in a box, so that no ball goes to a box of its own colour isA. `40`B. `44`C. `45`D. `60` |
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Answer» Correct Answer - B `(b)` `(D_(n))/(n!)-(D_(n-1))/((n-1)!)=((-1)^(n))/(n!)` gives `(D_(n))/(n!)=sum_(r=2)^(n)((D_(r ))/(r!)-(D_(r-1))/((r-1)!))=sum_(r=2)^(n)((-1)^(r ))/(r!)` `impliesD_(n)=n!sum_(r=2)^(n)((-1)^(r ))/(r!)` `:. D_(5)=5!((1)/(2!)-(1)/(3!)+(1)/(4!)-(1)/(5!))=44` |
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| 178. |
A person tries to form as many different parties as he can, out of his20 friends. Each party should consist of the same number. How many friendsshould be invited at a time? In how many of these parties would the samefriends be found? |
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Answer» Let the person invite r number of friends at a time. Then, the number of parties is `.^(20)C_(r )`, which is maximum when r=10. Ifa particular friend will be found in x parties, then x is the number of combinations out of 20 in which this particular friend must be included. Therefore, we have to select 9 more from 19 remaining friends. Hence, `x= .^(19)C_(9)`. |
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| 179. |
Find the number of ways of selecting 3 pairs from 8 distinct objects. |
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Answer» Number of ways of selecting first pair of objects `= .^(8)C_(2)` Number of ways of selecting second pair of objects `= .^(6)C_(2)` Number of ways of selecting third pair of objects `= .^(4)C_(2)` `therefore` Number of ways `=(.^(8)C_(2)xx .^(6)C_(2) xx .^(4)C_(2))//3!`, since the order in which pairs are selected is immaterial. |
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| 180. |
In a certain an algebraical exercise book there and 4 examples onarithmetical progression, 5 examples on permutation and combination, and 6examples on binomial theorem. Find the number of ways a teacher can select orhis pupils at least one but not more than 2 examples from each of these sets. |
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Answer» Number of ways teacher can select examples from arithmetic progression `=(.^(4)C_(1)+ .^(4)C_(2))` Number of ways teacher can select examples from permutation and combination `=(.^(5)C_(1) + .^(5)C_(2))` Number of ways teacher can select examples from binomial theorem `=(.^(6)C_(1)+ .^(6)C_(2))` Hence, total number of ways `=(.^(4)C_(1)+ .^(4)C_(2))(.^(5)C_(1)+ .^(5)C_(2))(.^(6)C_(1)+ .^(6)C_(2))` |
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| 181. |
Twenty-eight games were played in a football tournament with each teamplaying once against each other. How many teams were there? |
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Answer» Let number of teams be n. Number of matches to be played = number of ways two teams can be selected `= .^(n)C_(2)=28` `therefore (n(n-1))/(2)=28` `implies n^(2)-n-56=0` `implies (n-8)(n+7)=0` `implies n=8 " as" n ne -7`. |
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| 182. |
If a secretary and a joint secretary are to be selected from a committee of 11 members, then in how many ways can they be selected ?A. 110B. 55C. 22D. 11 |
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Answer» Correct Answer - B Selection of 2 members out of 11 has `.^(11)C_(2)` number of ways `.^(11)C_(2) = 55` |
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| 183. |
There are four chairs with two chairs in each row. In how many ways can four persons be seated on the chairs, so that no chair remains unoccupiedA. 6B. 12C. 24D. 48 |
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Answer» Correct Answer - C First chair can be occupied in 4 ways and second chair can be occupied in 3 ways, third chair can be occupied in 2 ways and last chair can be occupied in one ways only. So total number of ways `= 4 xx 3 xx 2 xx 1 = 24` |
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| 184. |
A bag contains 50 tickets numbered 1, ,2 3,...50. Find the number of set of five tickets `x_1 |
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Answer» Correct Answer - `.^(29)C_(2)xx .^(20)C_(2)` Since `x_(1) lt x_(2) lt x_(3) lt x_(4) lt x_(5) " and" x_(3)=30`, therefore, `x_(1),x_(2) lt 30`, i.e., `x_(1) " and" x_(2)` should come from tickets numbered from 1 to 29 and this may happen in `.^(29)C_(2)` ways. Now `x_(4),x_(5)` should come from 20 tickets numbered from 31 to 50 in `.^(20)C_(2)` ways. So, total number of ways is `.^(29)C_(2) .^(20)C_(2)`. |
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