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Solved Objective Question on Permutations and Combinations Set 2

Posted on - 17-05-2017

JEE Math PC

IIT JEE

Question.1

The number of zeroes at the end of (127)! is

(A) 31

(B) 30

(C) 0

(D) 10

Solution

Number of zeroes =

= 25 + 5 + 1 = 31

Question.2

The value of the expression k-1Ck-1 + kCk-1 + . . . + n+k-2Ck-1 is .

(A) n+ k-1Ck+1

(B) n+ k-1Ck-1

(C) n+ kCk

(D) none of these

Solution

We have k-1Ck-1 + kCk-1 + . . . + n+k-2Ck-1.

= kCk + kCk-1 + k+1Ck-1 . . . + n+k-2Ck-1 [ since kCk = 1 = k-1Ck-1].

= k+1Ck + k+1Ck-1 +. . . . + n+k-2Ck-1 [ since kCk + kCk-1= k+1Ck] .

= k+2Ck + . . . . + n+k-2Ck-1 = . . .

= n+k-2Ck + n+k-2Ck-1 =n+k-1Ck.

Question.3

The number of times of the digits 3 will be written when listing the integer from 1 to 1000 is

(A) 269

(B) 300

(B) 271

(D) 302

Solution

Since 3 does not occur in 1000, we have to count the number of times 3 ocures when we list the integers from 1 to 999. Any number between 1 and 999 is of the form xyz where 0 £ x, y, z£ 9 . Let us first count the numbers in which 3 occurs exactly once. Since 3 can occur at one place in 3C1 ways, there are 3C1 (9´9) = 3´ 92 such numbers. Next, 3 can occur exactly two places in (3C2) (9) = 3 ´ 9 such numbers .

Lastly, 3 can occur in all three digits in one number only. Hence the number of times 3 occurs is 1 ´ (3 ´ 92) + 2´(3´9) + 3´1 = 300.

Question.4

The number of natural numbers which are less than 2.108 and which can be written by means of the digits 1 and 2 is .

(A) 772

(B) 870

(C) 900

(D) 766

Solution

The required numbers are 1, 2, 11, 12, 21, 22, . . . ,122222222. .

Let us calculate how many numbers are these.

There are 2 one-digit such numbers. There are 22 two-digit such numbers and so on.

There are 28 eight-digit such numbers. All the digit numbers beginning with 1 and written by means of 1 and 2 are smaller than 2.108. Thus, there are 28 such nine-digit numbers. .

Hence the required number of numbers is

2 + 22 + 23+ …. + 28 + 28 = + 28 = 29 – 2 + 28 = 766.

Question.5

In a certain test, there are n questions. In this test 2n-i students gave wrong answers to at lest i question, where i = 1, 2, . . . , n .If the total number of wrong answers given is 2047, then n is equal to.

(A) 10

(B) 11

(C) 12

(D) 13

Solution

The number of students answering exactly i ( 1£ i £ n-1) questions wrongly is 2n-i – 2n-i-1. The number of students answering all n questions wrongly is 20 . Thus, the total number of wrong answers is .

1(2n-1 – 2n-2) +2( 2n-2 – 2n-3) + . . . + ( n-1)( 21 - 20) +n(20).

= 2n-1 + 2n-2 + 2n-3 + . . . + 20 = 2n – 1.

Thus 2n –1 = 2047
&⇒
2n = 2048 = 211
&⇒
n = 11 .

Question.6

Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. The number of ways in which we can place the balls in the boxes ( order is not considered in the box) so that no box remains empty is .

(A) 150

(B) 300

(C) 200

(D) none of these

Solution

One possible arrangement is

Three such arrangements are possible. Therefore, the number of ways is (5C2) (3C2) (1C1) (3) = 90 .

The other possible arrangements

Three such arrangements are possible. In this case, the number of ways is ( 5C1) ( 4C1) ( 3C3) (3) = 60 .

Hence, the total number of ways is 90 + 60 = 150.

Question.7

Let A = {x l x is a prime number and x < 30} .The number of different rational numbers whose numerator and denominator belong to A is .

(A) 90

(B) 180

(C) 91

(D) none of these

Solution

A = {2, 3, 5, 7, 11,13, 17, 19, 23, 29}. A rational number is made by taking any two in any order . So, the required number of rational numbers = 10P2 +1 (including 1).

Question.8

The number of ways of arranging six persons (having A, B, C and D among them) in a row so that A, B, C and D are always in order ABCD (not necessarily together) is

(A) 4

(B) 10

(C) 30

(D) 720

Solution

The number of ways of arranging ABCD is 4!. For each arrangement of ABCD, the number of ways of arranging six persons is same. Hence required number is

Question.9

Let A be the set of 4-digit numbers a1a2a3a4 where a1> a2> a3> a4, then n(A) is equal to

(A) 126

(B) 84

(C) 210

(D) none of these

Solution

Any selection of four digits from the ten digits 0, 1, 2, 3, . . . , 9 gives one such number. So, the required number of numbers = 10C4 = 210 .

Question.10

Let S be the set of all functions from the set A to the set A. If n(A) = k, then n(S) is .

(A) k!

(B) kk

(C) 2k – 1

(D) 2k

Solution

Each element of the set A can be given the image in the set A in k ways. So, the required number of functions, i.e. , n(S) = k´ k ´ . . .(k times) = kk . .

 
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