This EduRev document offers 20 Multiple Choice Questions (MCQs) from the topic Permutation & Combination (Level - 2). These questions are of Level - 2 difficulty and will assist you in the preparation of CAT & other MBA exams. You can practice/attempt these CAT Multiple Choice Questions (MCQs) and check the explanations for a better understanding of the topic.
Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:There are four rooms in a lodge: one single, one double, one for three persons and one for four persons. In how many ways can 10 persons be placed in these rooms?
Explanation
10 persons can be arranged in 10! ways.
if 1 person is placed in a room for single person = 10!/1!
If 2 persons are placed in room for double person, then the arrangements = 10!/2!
If 3 persons are placed in the room for three persons then number of arrangements = 10!/3!
If 4 persons are placed in a room for four people, then number of arrangements = 10!/4!
So, total number of arrangements = 10!/2!3!4!
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:There are 5 letters and 5 directed envelopes. The number of ways in which all the letters can be put in a wrong envelope are
Explanation
Number of ways in which 'n' objects can be placed on 'n' positions in such a manner that none of them is correct is given by the dearrangement formula.
Dearrangement (n) = n!(1/0! - 1/1! + 1/2! - 1/3!.... 1/n!)
Number of ways
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:A five-digit number is formed by the digits 1, 2, 3, 4 and 5 without repetition. Find the probability that the number is divisible by 4.
Explanation
Total number of five-digit numbers formed by the digits 1, 2, 3, 4 and 5 is 5!.
Therefore, exhaustive number of cases = 5! = 120
A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
Therefore, the last two digits can be 12, 24, 32 or 52. The last two digits can be filled in 4 ways.
Corresponding to each of these ways, there are 3! = 6 ways of filling the remaining three places.
∴ The total number of five-digit numbers formed by the digits 1, 2, 3, 4 and 5 and divisible by 4 is 4 × 6 = 24
Therefore, favourable number of cases = 24
So, required probability = 24/120 = 1/5
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:How many five-digit numbers can be formed that are the same when the order of their digits is reversed?
Explanation
We require that the 1st and the 5th numbers are the same and the 2nd and the 4th numbers are the same. If the 1st and the 2nd numbers are chosen, the 5th and the 4th numbers will also have been chosen. As the 1st number cannot be 0, it can be chosen in 9 ways. The 2nd and the 3rd numbers can be chosen in 10 ways each. Therefore, the total number of ways is (9 × 102) = 900.
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:A committee of four men and three women is to be formed from among six men and four women. If a particular woman, W1, refuses to be in the committee with another woman, W2, then in how many ways can the committee be formed?
Explanation
4 men can be chosen from among 6 in 6C4 ways and 3 women from among 4 can be chosen in 4C3 ways, i.e. 60 ways.
If 2 women, W1 and W2, are in the same committee, then the 3rd woman can be chosen in 2 ways.
In this case, the number of ways will be 6C4 × 2 ways.
Therefore, the number of ways of forming the committee if W1 and W2 are never in the same committee = 60 - 30 = 30
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:A code is of the format XXXXYYYY, where X can be a letter from A-H and Y can be a digit from 0-9 such that no repetition of any letter or digit occurs (e.g. ABHE0387). How many such codes exist?
Explanation
The code is formed by 4 letters and 4 digits. The 4 letters are first to be selected and then to be arranged, and the same has to be done for digits.
Number of ways to select and arrange the letters = 8C4 × 4!
Number of ways to select and arrange the digits = 10C4 × 4!
The code is formed by both letters and digits, so total number of cases = (8C4 × 4!) × (10C4 × 4!)
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:In an ice-cream store, one can pick from six flavours - Mango, Chocolate, Blueberry, Strawberry, Cherry and Vanilla. Further, one may also choose some topping from Choco-chip, Almond, Cashew, Pistachio and Coconut. One can even combine some or all of these toppings. How many different ice-creams are possible, if ice-cream of only one flavour can be chosen?
Explanation
Ice-creams with no topping = 6
Ice creams with one topping = 6 × 5 = 30
Ice-creams with two toppings = 6 × 5C2 = 60
Ice-creams with three toppings = 6 × 5C3 = 60
Ice-creams with four toppings = 6 × 5C4 = 30
Ice-creams with five toppings = 6 × 5C5 = 6
So, total number of different ice-creams possible = 6 + 30 + 60 + 60 + 30 + 6 = 192 = 6 × 25
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:
Then, n is
Explanation
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:A city having eight hotels gets five tourists. In how many ways can these tourists stay in these hotels, if at least two tourists have to stay together?
Explanation
The total number of ways in which five tourists can stay in eight hotels is (8)5. The number of ways in which no two tourists stay together = 8 × 7 × 6 × 5 × 4 = 6720. Therefore, the number of ways in which at least two tourists stay together = (8)5 – 6720 = 32,768 - 6720 = 26,048
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:How many five digit numbers contain exactly one 3?
Explanation
Since we require that there is exactly one 3, this digit could occupy any of the five places. If the number does not start with 3, the first digit must necessarily be a digit other than 3 and 0. In such a case, the first digit can be chosen in eight ways and the remaining digits can be chosen in 9 ways each. The numbers will be of the following forms: 3XXXX, X3XXX, XX3XX, XXX3X and XXXX3. Correspondingly, the total number of ways is (94) + (4 × 8 × 93) = 29889.
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:There are 3 piles of identical red, blue and green balls and each pile contains at least 10 balls. The number of ways of selecting 10 balls, if twice as many red balls as green balls are to be selected, is
Explanation
Let the number of green balls be x.
Then, the number of red balls is 2x. Let the number of blue balls be y.
Then, x + 2x + y = 10
⇒ 3x + y = 10
⇒ y = 10 - 3x
Clearly, x can take values 0, 1, 2 and 3. The corresponding values of y are 10, 7, 4 and 1.
Thus, the possibilities are (0, 10, 0), (2, 7, 1), (4, 4, 2) and (6, 1, 3), where (r, b, g) denotes the numbers of red, blue and green balls.
So, the answer is Option (2).
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:Seven distinct academic prizes are to be given to some students from a class of 50 students. Four prizes out of these are to be given to 4 distinct students. The rest three prizes can be given to any student even if he has won any other prize before. In how many ways can this be done?
Explanation
For the first 4 prizes, four students have to be selected. Number of ways to do this = 50C4.
Further, 3 prizes can be given to anyone. So, the possibility for each prize is 50.
Therefore, total number of possibilities = 50C4 × (50)3
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:The letters of the word RANDOM are written in all possible orders and the words so obtained are arranged in a dictionary order. In this list, the rank of the word RANDOM is
Explanation
Letters will occur in the order A, D, M, N, O and R.
The number of words which begin with A are 5P5 = 120.
Similarly, there are 120 words which begin with D, 120 words which begin with M, 120 words which begin with N and 120 words which begin with O.
So, 601st word begins with R.
There are 3P3 = 6 words which have RAD as the first three letters and 3P3 = 6 words which have RAM as the first three letters.
So, after 600 + 6 + 6 = 612 words, we have the words which have RAN as first three letters.
613rd word is RANDMO and 614th word is RANDOM.
So, the answer is option (2).
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:How many four-letter words can be formed if the first and the last letters are vowels (if repetition is allowed)?
Explanation
The first and the last letters can be filled in five ways each.
The remaining letters can be filled in 26 ways each.
Therefore, the total number of ways in which a four-letter word can be formed is 5 × 26 × 26 × 5 = 16,900.
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:In how many ways can two unit squares, 1 white and 1 black, be selected from an 8 x 8 chessboard such that they don't lie in the same row or column?
Explanation
White can be selected in 32 ways (because in a chessboard, there are 32 white and 32 black squares). There are 4 black squares in any row and 4 in any column. So, black can be selected in (32 - 8) = 24 ways.
Thus, total ways of selecting 1 white and 1 black = 32 x 24 = 768
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:In an eating competition, Mr. Bhima has to eat 10 different sweets (one at a time) named A to J. In how many different ways can he eat all the sweets, such that he has to eat A before B and B before C?
Explanation
If 'n' things are there out of which 'r' has to follow an order, then the total number of arrangements = n!/r!
So, different ways in which Mr. Bhima can eat all the sweets, such that he has to eat A before B and B before C = 10!/3! = 604,800
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:How many four-lettered palindromes (same in both the directions), with or without meaning can be formed using the English alphabet?
Explanation
4-lettered palindrome means the first and the last, and the 2nd and the 3rd letters should be the same.
The first letters can be any out of 26.
Similarly, the second letters also can be any out of 26.
So, the required number of palindromes is 26 × 26 = 262
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:There are 20 families with 4 members each, living in a 20-storey building. The families mutually decide to form a welfare committee for the well-being of all the residents. The committee will be of 10 members. Each family's only senior two members are eligible for membership in the committee and no more than 1 member from any family can be a part of the committee. In how many different ways can the committee be constituted?
Explanation
Ten members in the committee means 1 member each from 10 different families.
The number of ways to select 10 families from 20 families is 20C10.
Now, for each family, there are two different possibilities for the selection of members. Therefore, the number of ways in which this can be done for the 10 families is 210.
These two conditions co-exist. So, total number of cases = 20C10 × 210
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:How many 4-digit numbers divisible by 6 can be formed from the digits 1, 2, 3, 4, 5 and 6 without repetition?
Explanation
If the number is divisible by 6, it has to be divisible by both 2 and 3.
If it has to be divisible by 2, it has to be even.
If it is divisible by 3, then the sum of the digits has to be divisible by 3.
Possible 4-digit combinations are: (1, 2, 3, 6); (1, 2, 4, 5); (2, 3, 4, 6); (1, 3, 5, 6); (3, 4, 5, 6)
Number of cases for (1, 2, 3, 6) = 3 × 2 × 1 × 2 = 12
Number of cases for (1, 2, 4, 5) = 3 × 2 × 1 × 2 = 12
Number of cases for (2, 3, 4, 6) = 3 × 2 × 1 × 3 = 18
Number of cases for (1, 3, 5, 6) = 3 × 2 × 1 × 1 = 6
Number of cases for (3, 4, 5, 6) = 3 × 2 × 1 × 2 = 12
Total number of cases = 12 + 12 + 18 + 6 + 12 = 60
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Question for Practice Questions Level 2: Permutation & Combination - 2
Try yourself:How many natural numbers greater than 7,00,000 can be formed by using the digits 0, 3, 5, 6, 7, 8, if repetition is not allowed?
Explanation
First digit should be 7 or 8 as the number should be greater than
7,00,000, so number of ways arranging that is 2.
Ways of arranging rest of the 5 digits will be 5!.
So, required number of natural numbers = 2 × 5! = 240
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