men and women are to seated in a row so that no two women sit together. If then show that the number of ways in which they can be seated is...

men can be seated in a row in ways. Now there are places in which women can be arranged in ways. ...

(m) men and (n) women are to seated in a row so that no two women sit together. If (m n) then show that the number of ways in which they can be seated is (frac{underline{m}underline{m + 1}}{underline{m - n + 1}}).: Permutations and Combinations (Mathematics)

(m) men and (n) women are to seated in a row so that no two women sit together. If (m n) then show that the number of ways in which they can be seated is (frac{underline{m}underline{m + 1}}{underline{m - n + 1}}).: Permutations and Combinations (Mathematics) | DivineJEE

In a college of 300 students, every student reads 5 newspapers and every newspaper is read by 60 students. The number of newspaper is

For (2 leq r leq n,binom{n}{r} + 2binom{n}{r - 1} + binom{n}{r - 2} =)

A student is allowed to select at most n books from a collection of (2n + 1) books. If the total number of ways in which he can select at least one book is 63 , find the value of n.

In how many ways can a pack of 52 cards be (i) divided equally among four players in order (ii) divided into four groups of 13 cards each (iii) divided in 4 sets, three of them having 17 cards each and the fourth just one card?

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1,2 and 3 only, is

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 is

Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have atleast one letter repeated, is

A five digit number divisible by 3 is to be formed using the numerals 0,1,2,3,4 and 5 without repetition. The total number of ways this can be done is

The value of ^47C4 + sumj = 1^5mspace2mu ^52 - jC3 is equal to

Download Divine JEE App

Quick Links

Exam Categories

Contact Info

Divine JEE