# Permutation & Combination Questions

## Prove that value of zero factorial is 1.

Solution : We have, P(n, r) = $$n!\over (n – r)!$$ Putting r = n, $$\implies$$  P(n, n) = $$n!\over 0!$$ $$\implies$$  n! = $$n!\over 0!$$               [ $$\because$$  P(n, n) = n! ] $$\implies$$  0! = $$n!\over n!$$  = 1 Hence, Proved.

## The number of ways in which 6 men and 5 women can dine at a round table, if no two women are to sit together, is given by

Solution : The number of ways to n people on circular table is (n-1)! So, first we fix position of men, the number of ways to sit men = 5! Now, women can sit in the gaps between men, there are 6 gaps between 5 mens, So, women can sit in $$^6P_5$$ ways Hence, total …

## A student is to answer 10 out of 13 questions in an examination such that he must choose atleast 4 from the first five questions. The number of choices available to him is

Solution : The number of choices available to him = $$^5C_4$$ x $$^8C_6$$ + $$^5C_5$$ x$$^8C_5$$ = 5.4.7 + 8.7 = 140 + 56 = 196 Similar Questions The number of ways in which 6 men and 5 women can dine at a round table, if no two women are to sit together, is given …

## The number of ways of distributing 8 identical balls in 3 distinct boxes, so that none of the boxes is empty, is

Solution : The number of ways of distributing n identical things in m different boxes is $$^{n-1}C_{m-1}$$ So, Required number of ways = $$^{8-1}C_{3-1}$$ = $$^7C_2$$ = 21 Similar Questions A student is to answer 10 out of 13 questions in an examination such that he must choose atleast 4 from the first five questions. …

## How many ways are there to arrange the letters in the word ‘GARDEN’ with the vowels in alphabetical order ?

Solution : Total number of ways in which all the letters can be arranged in alphabetical order = 6! There are two vowels (A, E) in the word ‘GARDEN’. Total number of ways in which these two vowels can be arranged = 2! $$\therefore$$ Total number of required ways = $$6!\over 2!$$ = 360 Similar …

## If the letters of the word ‘SACHIN’ are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number

Solution : In the word SACHIN’ order of alphabets is A, C, H, I, N and S. Number of words start with A = 5!, so with C, H, I, N. Now, words start with S and after that ACHIN are in ascending order of position, so 5.5! = 600 words are in dictionary before, …

## At an election, a voter may vote for any number of candidates not greater than the number of candidates to be elected. There are 10 candidates and 4 are to be elected. If a voter votes for atleast 1 candidate, then the number of ways in which he can vote is

Solution : Total Number of ways = $$^{10}C_1$$ + $$^{10}C_2$$ + $$^{10}C_3$$ + $$^{10}C_4$$ = 10 + 45 + 120 + 210 = 385 Similar Questions A student is to answer 10 out of 13 questions in an examination such that he must choose atleast 4 from the first five questions. The number of choices …

## The set S = {1,2,3,…..,12} is to be partitioned into three sets A, B and C of equal size. Thus, $$A\cup B\cup C$$ = S $$A\cap B$$ = $$B\cap C$$ = $$A\cap C$$ = $$\phi$$ The number of ways to partition S is

Solution : first we choose 4 numbers from 12 numbers, then 4 from remaining 8 numbers, and then 4 from remaining 4 numbers So, Required number of ways = $$^{12}C_4$$ x $$^8C_4$$ x $$^4C_4$$ = $$12!\over (4!)^3$$ Similar Questions How many different words can be formed by jumbling the letters in the word ‘MISSISSIPPI’ in …

## How many different words can be formed by jumbling the letters in the word ‘MISSISSIPPI’ in which no two S are adjacent ?

Solution : Given word is MISSISSIPPI, Here, I occurs 4 times, S = 4 times P = 2 times, M = 1 time So, we write it like this _M_I_I_I_I_P_P_ Now, we see that spaces are the places for letter S, because no two S can be together So, we can place 4 S in …

## From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then, the number of such arrangements is

Solution : The number of ways in which 4 novels can be selected = $$^6C_4$$ = 15 The number of ways in which 1 dictionary can be selected = $$^3C_1$$ = 3 Now, we have 5 places in which middle place is fixed. $$\therefore$$  4 novels can be arranged in 4! ways $$\therefore$$  total number …