JEE Questions for Maths Permutations And Combinations Quiz 9 - MCQExams.com

⁶C₃ – 7 excluding (2,3,5) (2,3,6)(2,3,7)(2,4,6)(2,4,7)(2,5,7) (3,4,7).

Required number of ways = ¹²C₃ – ⁷C₃

                                      = 220 – 35 = 185.

The number of points of intersection of 37 straight lines is ³⁷C₂ . But 13 of them pass through the point A. Therefore, instead of getting  ¹³C₂  points we get merely one point.

Similarly 11 straight lines out of the given 37 straight lines intersect at B. Therefore, instead of getting  ¹¹C₂ points, we get only one points. Hence the number of intersection points of the lines is

³⁷C₂ – ¹³C₂  – ¹¹C₂ + 2 = 535.

Number of words in which all the 5 letters are repeated = 10⁵ = 100000 and the number of words in which no letter is repeated are ¹⁰P₅ = 30240.

Hence the required number of ways are

100000 – 30240  =  69760.

We know that a five digit number is divisible by 3, if and only if sum of its digits is divisible  by 3, therefore, we should not use 0 or  3 while forming the five digit numbers. Now,

1. In case we do not use 0 the five digit  number can be formed (from the digit 1,2,3,4,5) in  ⁵P₅ ways.
2. In case we do not  use 3, the five digit number can be formed (from the digit 0,1,2,4,5) in
3. P₅ – ⁴P₄ = 5! - 4! = 120 – 24  = 96 ways.

Therefore, the total number of such 5 digit number

= ⁵P₅ + (⁵P₅ – ⁴P₄) = 120 + 96 = 216.

Out of 10 persons, A is in and G and H are out of the team, so we have to select 4 more from 7 remaining . This can be done in ⁷C₄ ways.These 5 persons can be arranged in a line in 5! Ways . Hence the number of possible arrangements is ⁷C₄·5! = ⁷C₃ · (5!)

According to dictionary

Here we have 1 M, 4I , 4S and 2P.

Therefore total number of selections of one or more letters = (1 + 1) (4 + 1)(4 + 1)(2 + 1) – 1 = 149

Let there be n men participants. Then the number of games that the men play between themselves is 2· ⁿC₂ and the number of games that the men played with the women is 2·(2n).

Therefore,  2· ⁿC₂ – 2 · 2n = 66     [By hypothesis]

⇒ n² – 5n – 66 = 0 ⇒ n = 11

The number of words before the words CRICKET is 4 × 5! + 2 × 4! + 2! = 530.

Since the balls are to be arranged in a row so that the adjacent balls are of different colours, therefore we can begin with a white ball or a black ball. If we begin with a white ball, we find that (n + 1) white balls numbered 1 to (n + 1) can be arranged in a row in (n + 1)! ways. Now (n+1) places are created between n + 1 white balls which can be filled by (n + 1) black balls in (n + 1)! ways.

So the total number of arrangements in which adjacent balls are of different colours and first ball is a white ball is (n + 1)! × (n × 1)! = [(n + 1)!)².But we can begin with a black ball also. Hence the required number of arrangements is 2[(n + 1)!]² .

A number between 5000 and 10,000 can have any of the digits 5,6,7,8,9 at thousand’s place. So, thousand’s place can be filled in 5 ways . Remaining 3 places can be filled  by the remaining 8 digits in ⁸P₃ ways . Hence required number = 5 × ⁸P₃ .

The result is trivially true for r = 1,2. It can be easily proved by the principle of mathematical induction that the result is true for r also.

Seven boys can be seated in a row in 7! ways.Hence  the total number of arrangement such that no two girls seated together = 7! × ⁸P₃.

In all, we have 8 squares in which 6X’s have to be placed and it can be done in ⁸C₆ = 28 ways. But this includes the possibility that either the top horizontal row does not have any X or the bottom horizontal has no X. Since we want each row must have at least one X, these two possibilities are to be excluded. Hence required number of ways are 28 – 2 = 26.

Four first prizes can be given in 20⁴ ways since first prize of Mathematics can be given in 20ways, first prize of Physics also in 20 ways , similarly first prizes of  Chemistry and English can be given in 20 ways each. (Note that a boy can stand first in all the four subjects ).

Then two second prizes can be given in 19² ways since a boy cannot get both the first and second prizes.

Hence the required number of ways  = 20⁴ × 19².

By inspection n = 10.

We have 32 places for teeth. For each place we have two choices either there is a tooth or there is no tooth. Therefore, the number of ways to fill up these places is 2³² . As there is no person without a tooth, the maximum population is 2³² - 1.