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301.

The number of porper divisors of 1800 which are also divisible by 10, isA. 18B. 34C. 27D. none of these

Answer» We have `1800=2^(3)xx3^(2)xx5^(2)`
Clearly, the required number of proper divisors in same as the number of ways of selecting at least 2 and at least one 5 out 3 identical `2^(s),2` identical `3^(s)and2` identical `5^(s)`.
`:.` Required number of proper divisors `=3xx(2+1)xx2=18`.
302.

The number of even proper divisors of 5040, isA. 48B. 47C. 46D. none of these

Answer» We have, `5040=2^(4)xx3^(2)xx5xx7`
`:.` Number of even proper divisors
= Number of ways of selecting at least one, 2 and any number of `3^(s),5^(s)and7`
`=(4)xx(2+1)xx(1+1)xx(1+1)-1=47`.
303.

A five-digit number divisible by 3 is to be formedusing the digits 0, 1, 2, 3, 4, and 5, without repetition. The total numberof ways this can done isA. 216B. 600C. 240D. 3125

Answer» Correct Answer - A
we know that, a number is divisible by 3, when sum of digits in the number must be divisible by 3.
So, if we consider the digits 0, 1, 2, 4, 5, then ( 0 + 1 + 2 + 4 + 5) = 12)
We see that, sum is divisible by 3. Therefore, five-digit numbers using the digit
`0, 1, 2, 4, 5 = 4 xx 4 xx 3 xx 2 xx 1 = 96`
`{:(4, 4, 3, 2, 1):}`
and if we consider the digit 1, 2, 3, 4, 5, then `(1 + 2 + 3 + 4 + 5 = 15)`
This sum is also divisible by 3.
So, five - digit number can be formed using the digit 1, 2, 3, 4, 5 is `5"!"` ways.
Total number of ways `= 96 + 5"!" = 96 + 120 = 216`
304.

Statement-1: A 5-digit number divisible by 3 is to be formed using the digits 0,1,2,3,4,5 without repetition, then the total number of ways this can be done is 216. Statement-2: A number is divisible by 3, if sum of its digits is divisible by 3.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true

Answer» Correct Answer - A
We know that a number is divisible by 3, if the sum of its digits is divisible by 3. now, out of 0,1,2,3,4,5,6 if we take 1,2,4,5,6 or 1,2,3,4,5 or 0,3,4,5,6 or 0,2,3,4,6 or 0,1,3,5,6 or 0,1,2,4,5 or 0,1,2,3,6
`therefore`Total number of ways`=2xx.^(5)P_(5)+5xx(.^(5)P_(5)-.^(4)P_(4))`
`=240+480`
`=720`
Statement-1 is false, statement-2 is true.
305.

The numberof 5 digit numbers which are divisible by 4, with digits from the set `{1, 2, 3, 4, 5}`and the repetition of digits is allowed, is ________.

Answer» Correct Answer - 625
A number is divisible by 4 if last 2 digit number is divisible by 4.
`therefore` Last two digit number divisible by 4 from (1, 2, 3, 4, 5) are 12, 24, 32, 44, 52
`therefore` The number of 5 digit number which are divisible by 4, from the digit (1, 2, 3, 4, 5) and digit is repeated is
`5 xx 5 xx 5 xx (5 xx 1) = 625`
306.

If a seven-digit number made up of all distinctdigits 8, 7, 6, 4, 3, `xa n dy`divisibleby 3, thena. Maximum value of `b . c. x-y d.`e.is 9f. Maximum value of `g . h. x+y i.`j.is 12k. Minimum value of `l . m. x y n.`o.is 0p. Minimum value of `q . r. x+y s.`t.is 3

Answer» Correct Answer - 8
The sum of digits is divisible by 3.
i.e., `8+7+6+4+2+x+y` or 27+x+y is divisible by 3 `thereforex+y` must be divisible by 3.
then, possible ordered pairs are
(0,3),(3,0),(1,5),(5,1)(,0,9),(9,0),(3,9),(9,3)
`therefore`Number of ordered pairs=8
307.

How many different seven digit numbers are there the sum of whose digits is even ?

Answer» Let us consider 10 successive 7-digit numbers.
`a_(1)a_(2)a_(3)a_(4)a_(5)a_(6)" "0,`
`a_(1)a_(2)a_(3)a_(4)a_(5)a_(6)" "1`,
`a_(1)a_(2)a_(3)a_(4)a_(5)a_(6)" "2`,
. . . . . . . . .
`a_(1)a_(2)a_(3)a_(4)a_(5)a_(6)" "9`,
where `a_(1),a_(2),a_(3),a_(4),a_(5) and a_(6)` are some digits, we see that half of these 10 numbers, i.e., 5 numbers have an even sum of digits.
the first digit `a_(1)` can assume 9 different values and each of the digits `a_(2),a_(3),a_(4),a_(5) and a_(6)` can assume 10 different values.
the last digit `a_(7)` can assume only 5 different values of which the sum of all digits is even.
`therefore` There are `9xx10^(5)xx5=45xx10^(5)`, 7-digit numbers the sum of whose digits is even.
308.

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2and 3 only, isA. 55B. 66C. 77D. 88

Answer» A seven digit number using 1, 2, and 3 and having 10 as the of digits can be formed either by using digits 1, 1, 1,1,1,2,3or,1,1,1,1,2,2,2.
So, required number of seven digit numbers
`=(7!)/(5!)+(7!)/(4!3!)=42+35=77`
309.

The number of all five digit numbers which are divisible by 4 that can be formed from the digits 0,1,2,3,4 (without repetition) isA. 36B. 30C. 34D. None of these

Answer» Correct Answer - B
310.

An n-digit numberis a positive number with exactly`n`digits. Ninehundred distinct n-digit numbers are to be formed using only the three digits2, 5, and 7. The smallest value of `n`for whichthis is possible isa.`6`b. `7`c. `8`d. `9`A. 6B. 7C. 8D. 9

Answer» Correct Answer - B
Distinct n - digit numbers which can be formed using digits 2, 5 and 7 are `3^(n)`. We have to find n, so that
`3^(n) ge 900`
`rArr" " 3^(n - 2) ge 100`
`rArr" " n - 2 ge 5`
`rArr" " n ge 7`, so the least value of n is 7.
311.

The number of 4-digit numbers that can be made with the digits 1,2,3,4 and 5 in which atleast two digits are identical, isA. `4^(5)-5!`B. `505`C. 600D. 120

Answer» Correct Answer - B
312.

The number of different seven digit numbers that can be written using only three digits 1, 2 &3 under the condition that the digit 2 occurs exactly twice in each number.A. `""^(7)P_(5).2^(5)`B. `""^(7)C_(2).2^(5)`C. `""^(7)C_(2).5^(2)`D. none of these

Answer» Choose any two of the seven digits (in the seven digit number). This may be done in `""^(7)C_(2)` ways. Put 2 in these two digits. The remaining 5 digits may be arranged using 1 and 3 in `2^(5)` ways.
So, required number of numbers = `""^(7)C_(2)xx2^(5)`.
313.

The number of all possible selections of one or more questions from 10 given questions, each question having an alternative is :A. 310B. `2^(10)-1`C. `3^(10)-1`D. `2^(10)`

Answer» Correct Answer - C
314.

A polygon has 44 diagonals. The number of its sides areA. 11B. 12C. 13D. 15

Answer» Correct Answer - A
Let the convex polygon has n sides.
`therefore` Number of diagonals `= ""^(n)C_(2) - n`
According to the question,
`""^(n)C_(2) - n = 44`
`(n"!")/(2"!"(n -2)"!") - n = 44`
`rArr` `(n(n-1))/(2) - n = 44`
`rArr` `n[((n-1))/(2)-1]=44 " "rArr n((n-1-2)/(2)) = 44`
`rArr` `n(n-3) = 44 xx 2 " "rArr n(n-3) = 88`
`rArr` `n^(2) - 3n - 88 = 0 " "rArr (n-11) (n+8) = 0`
`rArr` `n = 11, - 8" "` `[because n = -8]`
`therefore` n = 11
315.

The total number of 4 digit numbers in which the digit are in descending order, isA. `""^(10)C_(4)xx4!`B. `""^(10)C_(4)`C. `(10!)/(4!)`D. none of these

Answer» The total number of arrangements of 10 digits 0, 1, 2,……,9 by taking 4 at a time `=""^(10)C_(4)xx4!`
We observe that in every arrangement in which the digits are in descending order. Hence,
The required number of numbers `=(""^(10)C_(4)xx4!)/(4!)=""^(10)C_(4)`.
316.

The number of `6` digit numbers that can be made with the digits `1, 2, 3` and `4` and having exactly two pairs of digits, isA. 480B. 540C. 1080D. none of these

Answer» The number of 6-digits has two pairs of digits and 2 different digits.
The number of ways of selecting 2 digits for two pairs and 2 other digits `=""^(4)C_(2)xx""^(2)C_(2)`. Now, we arrange these six digits which consists of two pairs and 2 different digits. This can be done in `(6!)/(2!2!)` ways.
`:.` Required numbers `=""^(4)C_(2)xx""^(2)C_(2)xx(6!)/(2!2!)=1080`.
317.

A polygon has 44 diagonals. The number of its sides areA. 9B. 10C. 11D. 12

Answer» Correct Answer - C
318.

No. of diagonals of a polygon are 170. No. of sides in this polygon are:A. 12B. 17C. 20D. 25

Answer» Correct Answer - C
319.

The number of words of four letters containingequal number of vowels and consonants, where repetition is allowed, isa. `105^2`b. `210xx243`c. `105xx243`d. `150xx21^2`A. `105^(2)`B. `210xx243`C. `105xx243`D. none of these

Answer» A for letter word containing equal number of vowels and consonants may contain one vowel and one consonant of four letter words containing one vowel and one consonant
`=""^(5)C_(1)xx""^(21)C_(1)xx(4!)/(2!2!)`
The number of four letter words containing two vowels and two consonants `=""^(5)C_(1)xx""^(21)C_(2)xx4!`
Hence, required number of words
`=""^(5)C_(1)xx""^(21)C_(1)xx(4!)/(2!2!)+""^(5)C_(2)xx""^(21)C_(2)xx4!` `=105xx6+10xx210xx24=210(3+240)=210xx243`
320.

How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?

Answer» vowels= I,O,U,E
consonants = N, V,L,T
total words`= (.^4C_3 .^4C_2) .^5P_5`
`= (4!)/(1!*3!)*(4!)/(2!*2!)*(5!)/(5-5)!`
`= 4 * (4*3*2)/4 * 5!`
`= 24*120`
`=2880`
answer
321.

A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?

Answer» `M_1,M_2,W_1,W_2,W_3`
total ways `= .^5C_3 = (5!)/(3!2!)`
`= (5*4*3!)/(3!2*1) = 10`
with 1 man & 2 women
`= .^2C_1.^3C_1`
`= (2!)/(1!1!) * (3!)/(2!1!)`
`=2*3= 6`
answer
322.

From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?

Answer» no of person `= 8`
as one person cant hold one than more position so
total `= 8*7= 56`
323.

All the letters of the word EAMCET are arranged in all possible ways. The number such arrangements in which no two vowels are adjacent to each other isA. 54B. 72C. 114D. 360

Answer» Correct Answer - B
324.

In how many ways can the letters of the word PERMUTATIONS be arranged if the(i) words start with P and end with S,(ii) vowels are all together,(iii) there are always 4 letters between P and S?

Answer» (i) `P ----------S`
`=(10!)/(2!)`
(ii) `--------`
`= (8!)/(2!) * 5!`
(iii) ` P ---- S`
`(.^10P_4*7!*2)/(2!)`
`= .^10P_4*7!`
325.

Ten IIT students and 2 DCE students sit in a row. The number of ways in which exactly 3 IIT students sit in between 2 DCE students

Answer» Total possible values=`10C_3*2!*3!*8!`
`8*2*10!`
`16*10!`.
326.

The number of ways the letters of the word PERSON cann be placed in the squares of the figure shown so that no row remain empty is A. `24xx6!`B. `26xx6!`C. `26xx7!`D. `27xx7!`

Answer» Correct Answer - B
327.

How many different signals can be given using any number of flags from 4 flags of different colours?

Answer» The signals can be made by using one or more flags at a time. Hence, by the fundamental principal of addition, the total number of signals
`=.^(4)P_(1)+.^(4)P_(2)+.^(4)P_(3)+.^(4)P_(4)`
`=4+(4xx3)+(4xx3xx2)+(4xx3xx2xx1)`
`=4+12+24+24=64`
328.

A is a set containing n elements. A subset P of A is chosen. The set A is reconstructed by replacing the elements of P. A subset of A is again chosen. Find the number of ways of choosing P and Q, so that (i) `P capQ` contains exactly r elements. (ii) `PcapQ` contains exactly 2 elements. (iii) `P cap Q=phi`

Answer» Let `A={a_(1),a_(2),a_(3), . . .,a_(n)}`
(i) The r elements in P and Q such that `PcapQ` can be chosen out of n is `.^(n)C_(R)` ways a general element of A must satisfy one of the following possibilities [here, general element be `a_(i)(1leilen)]`
(i) `a_(i)inP and a_(i) in Q`
(ii) `a_(i) in P and a_(i) cancel(in)Q`
(iii) `a_(i) in P and a_(i) in Q`
(iv) `a_(i) cancel(in)P and a_(i) cancel(in)Q`
Let `a_(1),a_(2), . . ,a_(r) in P capQ`
There is only one choice each of them (i.e., (i) choice) and three choices (ii), (iii) and (iv) for each of remaining (n-r) elements.
Hence, number of ways of remaining elements=`3^(n-r)`
Hence, number of ways in which `P capQ` contains
exactly r elements`=^(n)C_(r)xx3^(n-r)`
(ii) Put r=2, then `.^(n)C_(2)xx3^(n-2)`
(iii) Put r=0, then `.^(n)C_(0)xx3^(n)=3^(n)`.
329.

Number of words of 4 letters that can be formed with the letters of the word IIT JEE, isA. 42B. 82C. 102D. 142

Answer» Correct Answer - C
There are 6y letters I,I,E,E,T,J
The following cases arise:
Case I All letters are different
`.^(4)C_(4)=4!=24`
Case II Two alike and two different
`.^(2)C_(1)xx.^(3)C_(2)xx(4!)/(2!)=72`
Case III Two alike of one kind and two alike of another kind
`.^(2)C_(2)xx(4!)/(2!2!)=6`
Hence, number of words=24+72+6=102
330.

In how many ways can a cricket team of eleven players be chosen out a batch 15 players, if (i) a particular is always chosen. (ii) a particular player is never chosen?

Answer» (i) Since, particular player is always chosen. It means that 11-1=10 players are selected out of the
remaining 15-1=14 players,
`therefore`Required of ways`=.^(14)C_(10)=.^(14)C_(4)`
`=(14*13*12*11)/(1*2*3*4)=1001`
(ii) Since, particular player is never chosen. it means that 11 players are seletected out the remaining 15-1=14. players.
`therefore`Required number of ways=`.^(14)C_(11)=.^(14)C_(3)`
`=(14*13*12)/(1*2*3)=364`
331.

The number of ways to rearrange the letters of the word CHEESE isA. 120B. 240C. 720D. 6

Answer» Correct Answer - A
332.

Prove that `1+1* ""^(1)P_(1)+2* ""^(2)P_(2)+3* ""^(3)P_(3) + … +n* ""^(n)P_(n)=""^(n+1)P_(n+1).`

Answer» `LHS=.^(1)P_(1)+2*.^(2)P_(2)+3*.^(3)P_(3)+ . . .+n*.^(n)P_(n)`
`=underset(r=1)overset(n)(sum)r*.^(r)P_(r)=underset(r=1)overset(n)(sum){(r+1)-1}*.^(r)P_(r)`
`=underset(r=1)overset(n)(sum){(r+1)*.^(r)P_(r)-.^(r)P_(r))}`
`=underset(r=1)overset(n)(sum)(.^(r+1)P_(r+1)-.^(r)P_(r))` [from note (iii)]
`=.^(n+1)P_(n+1)-.^(1)P_(1)=.^(n+1)P_(n+1)-1`
`=RHS`
333.

Find the sum of all five digit numbers ,that can be formed using the digits `1, 2, 3, 4 and 5` (repetition of digits not allowed)

Answer» Required sum`=(5-1)!(1+2+3+4+5+((10^(5)-1)/(9))`
`=14*15*11111=39999960`
334.

If `(""^(n+2)C_(6))/(""^(n-2)P_(2))=11`, then n satifes the equationA. `n^(2)+n-110=0`B. `n^(2)+5n-84=0`C. `n^(2)+3n-108=0`D. `n^(2)+2n=0`

Answer» We have,
`(""^(n+2)C_(6))/(""^(n-2)P_(2))=11`
`implies((n+2)!)/((n-4)!6!)xx((n-4))/((n-2)!)=11`
`implies(n+2)(n+1)(n)(n-1)=11xx6!`
`implies(n+2)(n+1)(n)(n-1)=11xx10xx9xx8`
`impliesn+2=11impliesn=9`
This value of n satisfies the equation `n^(2)+3n-108=0`.
335.

Find the value of `r,` if (i) `.^(11)P_(r)=990` (ii) `.^(8)P_(5)+5*.^(8)P_(4)=.^(9)P_(r)`

Answer» (i) `because.^(11)P_(r)=990=11xx10xx9=.^(11)P_(3)`
`thereforer=3`
(ii) `because .^(8)P_(5)+5*.^(8)P_(4)=.^(9)P_(r)`
`implies.^(8)P_(4)((.^(8)P_(5))/(.^(8)P_(4))+5)=.^(9)P_(r)`
`implies.^(8)P_(4)(8-5+1+5)=.^(9)P_(r)` [ from note (v)]
`implies9*.^(8)P_(4)=.^(9)P_(r)`
`implies.^(9)P_(5)=.^(9)P_(r)` [from note (iii)]
`thereforer=5`
336.

If ` ^(k+5)P_(k+1)=(11(k-1))/2dot ^(k+3)P_(k )P_k`then the values of `k`are`7 a n d 11`b. `6 a n d 7`c. `2 a n d 11`d. `2 a n d 6`A. 7 and 11B. 6 and 7C. 2 and 11D. 2 and 6

Answer» Correct Answer - B
337.

Find the sum of all five digit numbers ,that can be formed using the digits `1, 2, 3, 4 and 5` (repetition of digits not allowed)A. 366000B. 660000C. 360000D. 3999960

Answer» We know that the sum of all n-digit numbers formed by using n digits from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 is
Hence, required sum `=(1+2+3+4+5)xx4!xx((10^(5)-1)/(10-1))`
`=360(100000-1)/(9)=3999960`
338.

Find n, ` ""^(n+5)P_(n+1)=(11)/(2)(n-1)*""^(n+3)P_(n).`

Answer» We have, `(.^(n+5)P_(n+1))/(.^(n+3)P_(n))=(11(n-1))/(2)`
`implies((n+5)(n+3)^(n+3)P_(n-1))/(.^(n+3)P_(n))=(11(n-1))/(2)` [from note (iii)]
`((n+5)(n+4))/((n+3-n+1))=(11(n-1))/(2)` [from note (v)]
`implies(n+5)(n+4)=22(n-1)`
`impliesn^(2)-13n+42=0`
`implies(n-6)(n-7)=0`
`thereforen=6.7`
339.

If `""^(12)P_(r)=""^(11)P_(6).""^(11)P_(5)` then r is equal toA. 6B. 5C. 7D. none of these

Answer» Correct Answer - A
340.

The number of signals that can be generated by using 6 differently coloured flags, when any number of them may be hoisted at a time isA. 1956B. 1957C. 1958D. 1959

Answer» The singals can be made by using at a time one or more flags.
The total number of singals when r flags are used at a time from 6 flags is equal to the number of arrangements of 6, taking r at a time i.e., `""^(6)C_(r)xxr!`. Since r can take values 1, 2, 3, 4, 5, 6. Hence, the number of different signals that can be generated is
`sum_(r=1)^(6)""^(6)C_(r)xxr!""=6+30+120+360+720+720=1956`
341.

Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available.

Answer» for 2 flags `5*4 = 20`
for 3 flags`5*4*3= 60`
for 4 flags `5*4*3*2 = 120`
for 5 flags `5*4*3*2*1 = 120`
total ways= `20+60+120+120`
`=320`
answer
342.

The number of divisors of the form `(4n+2)` of the integer 240 isA. 4B. 8C. 10D. 3

Answer» We have,
`4m+2and240=2^(4)xx3xx5`
Therefore, in any divisor of the form 4m+2, number 2 must occur exactly once. Hence, the number of divisors of the form 4m+2 is
`1xx(1+1)(1+1)=4`
343.

What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these(i)         four cards are of the same suit,(ii)        four cards belong to four different suits,(iii)       are face cards,(iv)       two are red cards and two are black cards,(v)        cards are of the same colour?

Answer» (i) `4.^13C_4`
(ii) `.^13C_1.^13C_1.^13C_1.^13C_1`
`= 13^4`
(iii) `.^12C_4`
(iv) `.^26C_2.^26C_2`
(v) `.^26C_4*2`
answer
344.

The number of ways in which n distinct objects can be put into two identical boxes so that no box remains empty, isA. `2^(n)-2`B. `2^(n)-1`C. `2^(n-1)-1`D. `2^(n)-1`

Answer» Let us first label the boxes `B_(1)andB_(2)`. Now each object can be put either in `B_(1)or"in" B_(2)`. So, there are two ways to deal with each of the n objects. Consequently, n objects can be dealt with `2^(n)` ways. Out of these are `2^(n)` ways (i) when all objects are put in box `B_(1)` (ii) when all objects are put in box `B_(2)`. Thus, there `2^(n)-2` ways in which neither box is empty. If we now remove the labels from the boxes so that they become identical, this number must be divided by bty 2 to get the required number of ways.
`:.` Required number of ways `=(1)/(2)(2^(n)-2)=2^(n-1)-1`
345.

Determine `n` if (i) `""^(2n)C_(3) : ""^(n)C_(3) = 12:1` (ii) `""^(2n)C_(3): ""^(n)C_(3)= 11:1`

Answer» (i) `C(2n,2):C(n,2) = ((2n!)/((2!)(2n-2)!))/((n!)/((2!)(n-2)!)`
`=(((2n)**(2n-1)**(2n-2)!)/((2!)(2n-2)!))/(((n)**(n-1)**(n-2)!)/((2!)(n-2)!))`
`=(2(2n-1))/(n-1)`
Now, we are given,`C(2n,2):C(n,2) = 12:1`
`:. (2(2n-1))/(n-1) = 12`
`=>4n-2 = 12n-12=> 8n = 10 => n = 5/4`
But, `n` should be a whole number.
So, there is no value of `n` that can solve this.

(ii) `C(2n,3):C(n,3) = ((2n!)/((3!)(2n-3)!))/((n!)/((3!)(n-3)!)`
`=(((2n)**(2n-1)**(2n-2)**(2n-3)!)/((2!)(2n-3)!))/(((n)**(n-1)**(n-2)**(n-3)!)/((3!)(n-3)!))`
`=(2(2n-1)(2n-2))/((n-1)(n-2))`
`=(4(2n-1))/(n-2)`
Now, we are given,`C(2n,3):C(n,3) = 11:1`
`:. (4(2n-1))/(n-2) = 11/1`
`=>8n-4 = 11n-22`
`=> 3n = 18=> n = 6`
346.

Consider three boxes, each containing 10 balls labelled 1, 2, .., 10. Suppose one ball is randomly drawn from each of the boxes. Denote by `n_(i)`, the label of the ball drawn from the ith box, (I = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that `n_(1) lt n_(2) lt n_(3)` isA. 82B. 120C. 240D. 164

Answer» Correct Answer - B
Given there are three boxes, each containing 10 balls labelled 1, 2, 3, .. , 10.
Now, one ball is randomly drawn from each boxes, and `n_(i)` denote the label of the ball drawn from the ith box, (i = 1, 2, 3).
Then, the number of ways in which the balls can be chosen such that `n_(1) lt n_(2) lt n_(3)` is same as selection of 3 different numbers from numbers `{1, 2, 3, .. , 10} = ""^(10)C_(3) = 120.`
347.

if `.^(2n)C_(2):^(n)C_(2)=9:2 and .^(n)C_(r)=10`, then r is equal toA. 2B. 3C. 4D. 5

Answer» Correct Answer - A
348.

If `.^(20)C_(n+1)=.^(n)C_(16)`, the value of n isA. 7B. 10C. 13D. None of these

Answer» Correct Answer - D
349.

If `""^(47)C_(4)+sum_(r=1)^(5) ""^(52-r)C_(3)` is equal toA. `.^(47)C_(6)`B. `.^(52)C_(5)`C. `.^(52)C_(4)`D. None of these

Answer» Correct Answer - C
350.

Consider all possible permutations of the letters of the word ENDEANOEL. Match the statements/ expressions in column I with the statement/expressions in

Answer» Correct Answer - A::B::C::D
`(A)to(p),(B)to(s),(C)to(q),(D)to(q)`
(A) ENDEA, N,O,E,L are five different letters, then perrmutations=5!.
(B) if E is in the first and last position, then permutations
`=(7!)/(2!)=(7xx6xx5!)/(2)=21xx5!`
(C) for first four letters`=(4!)/(2!)=4xx3=12` and for last five
letters`=(5!)/(3!)=(5!)/(6)`, then permutations `=12xx(4!)/(2!)=12`, then permutations `=(5!)/(3!)xx12=(5!)/(6)xx12=2xx5!`