Permutation & Combination Questions

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.

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 …

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 Read More »

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 …

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 Read More »

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. …

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

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 …

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

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, …

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 Read More »

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 …

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 Read More »

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 …

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 Read More »

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 …

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

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 …

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 Read More »