This EduRev document offers 10 Multiple Choice Questions (MCQs) from the topic Permutation & Combination (Level - 3). These questions are of Level - 3 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 3: Permutation & Combination - 1
Try yourself:How many ways are there to pick two different cards from a deck of 52 cards such that the first card is
(a) an ace and the second is not a queen
(b) a spade and the second is not a queen
Explanation
(a) First card, i.e. an ace can be drawn in 4 ways. Second card i.e. not a queen can be drawn in 51 – 4 i.e. 47 ways.
Therefore, the total number of ways = 4 × 47, i.e. 188 ways
(b) Case 1: Let the first card be a spade queen. Total number of ways = 1 × 48 = 48
Case 2: Let the first card be a spade card but not a queen. Total number of ways = 12 × 47 = 564
Therefore, the answer is 48 + 564 = 612.
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Question for Practice Questions Level 3: Permutation & Combination - 1
Try yourself:How many 4-digit numbers can be formed by using the digits 0, 2, 3, 5 and 8 (without repetition), such that each number is not divisible by 4?
Explanation
Total 4-digit numbers that can be formed = 4 × 4 × 3 × 2 = 96
If a number is divisible by 4, then the number formed by the last 2 digits should be divisible by 4.
So, 08, 20, 28, 32, 52 or 80 should be the last 2 digits.
If the last 2 digits are either 20, 80 or 08, the remaining two digits can be filled in 3 × 2 = 6 ways. Therefore, 6 x 3 = 18 ways.
If the last 2 digits are 28, 32 or 52, the remaining two digits can be filled in 4 ways. Therefore, 4 x 3 = 12 ways.
So, the required number of 4-digit numbers = 96 - (18 + 12) = 96 - 30 = 66
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Question for Practice Questions Level 3: Permutation & Combination - 1
Try yourself:In how many ways can 4 cards be drawn randomly from a pack of 52 cards such that there are at least 2 kings and at least 1 queen among them?
Explanation
The different possibilities are:
2 kings and 1 queen and 1 other card
Or, 3 kings and 1 queen
Or, 2 kings and 2 queens
Total ways possible = 4C2 × 4C1 × 44C1 + 4C3 × 4C1 + 4C2 × 4C2
= 1056 + 16 + 36
= 1108
Total ways possible = 1108
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Question for Practice Questions Level 3: Permutation & Combination - 1
Try yourself:The manager of a football team of eleven players wants to take photographs of his team, six players at a time. He can choose any one out of four renowned photographers to take the photographs. In how many ways can the photographs be taken, if the Captain and the Vice-Captain of the team are always included in the photographs? Also, find out the number of ways in which the photographs can be taken, if the Captain and the Vice-Captain of the team are never included in the photographs.
Explanation
If the Captain and the Vice-Captain are always included, the number of ways is 9C4 × 6!. But, these photographs can be taken by the photographers in 4 different ways. So, the required number of ways is 9C4 × 6! × 4.
If the Captain and the Vice-Captain are never included, the number of ways is 9C6 × 6!. But, these photographs can be taken by the photographers in 4 different ways. So, the required number of ways is 9C6 × 6! × 4 = 9P6 × 4.
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Question for Practice Questions Level 3: Permutation & Combination - 1
Try yourself:A 6-character code is made of only 0s and 1s, such that the number of 0's is always greater than the number of 1's. How many such codes exist?
Explanation
If the number of 0's is greater than the number of 1's, then the cases are:
i. Six 0's and no 1's. There is only one possibility that the code is 000000. So, number of codes = 1
ii. Five 0s and one 1s. The number of possibilities is: 6!/5! = 6 (5! is divided as 0 repeats 5 times)
iii. Four 0s and two 1s. The number of possibilities is: = 15 (4! is divided as 0 repeats 4 times and 2! is divided as 1 repeats two times)
Therefore, total number of codes = 1 + 6 + 15 = 22
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Question for Practice Questions Level 3: Permutation & Combination - 1
Try yourself:A club has 15 members; 12 of them being married couples and 3 being unmarried. These members have decided to form a leadership board in the club which consists of 4 members: President, Vice-president, Public Relations Officer and General Manager. The Public Relations Officer is to be selected from the unmarried members. For all the other positions, the member selected has to be married. However, one couple decides that they both together will not be a part of the leadership board. In how many ways can the board be formed?
Explanation
For the selection of the Public Relations Officer, the number of possibilities is 3.
Now, for the selection of the other members, cases have to be considered for the one couple which together won't be a part of the board.
If the husband is a part, from the 10 remaining married members, two more have to be selected and then positions have to be given to all three selected. Number of cases = 10C2 × 3!
If the wife is a part, Number of cases = 10C2 × 3!
If none of them is a part = 10C3 × 3!
Total = 3 × (2 × 10C2 × 3! + 10C3 × 3!)
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Question for Practice Questions Level 3: Permutation & Combination - 1
Try yourself:Six cards are to be selected from a deck of cards. At least 1 card is to be selected from each suit. In how many different ways can this be done?
Explanation
The possible card distribution for different suits:
(1, 1, 1, 3); (1, 1, 2, 2)
Number of cases for (1, 1, 1, 3) = 4C1 × 13C3 × (13C1)3 (4C1 is done to select the suit from which three cards will be selected. Let's say this is clubs. Now, 3 cards from clubs are to be selected, which can be done in 13C3 ways.
Further, 1 card each is to be selected from all the other suits. This can be done in 13C1 ways for each of the other suits.)
Number of cases for (1, 1, 2, 2) = 4C2 × (13C2)2 × (13C1)2
(Two suits are to be selected from the four suits, this can be done in 4C2 ways and further selections can be done like the above case.)
Total number of cases = 4C1 × 13C3 × (13C1)3 + 4C2 × (13C2)2 × (13C1)2
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Question for Practice Questions Level 3: Permutation & Combination - 1
Try yourself:Two groups are to be formed from 10 different boys of a class. The groups may or may not be of the same size. In how many different ways can this be done?
Explanation
The groups can be: (1, 9); (2, 8); (3, 7); (4, 6); (5, 5).
Number of ways to form group (1, 9) = 10C1
Number of ways to form group (2, 8) = 10C2
Number of ways to form group (3, 7) = 10C3
Number of ways to form group (4, 6) = 10C4
Number of ways to form group (5, 5) = 10C5 ÷ 2 (Suppose A, B, C, D, E, F, G, H, I and J are the names of the 10 boys. In selecting 5 of them, let's say A, B, C, H and J are selected, the other 5 left are D, E, F, G and I. So, we have two groups of 5 each. Now, if D, E, F, G and I are selected, then the other left boys are A, B, C, H and J. These two groups are the same group as formed above. So, we need to divide 10C5 by 2, because the groups formed are identical.)
Total number of ways = 10C1 + 10C2 + 10C3 + 10C4 + (10C5 ÷ 2)
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Question for Practice Questions Level 3: Permutation & Combination - 1
Try yourself:Some tennis balls are to be put into 7 boxes such that each box contains at least one ball. At the most, 3 of the boxes are to contain the same number of balls, and no two of the remaining boxes are to contain an equal number of balls. What is the least possible number of tennis balls needed for the boxes?
Explanation
Since at the most, 3 of the boxes are to contain the same number of balls, to minimise the number of balls in each box, suppose each box contains just 1 ball. Next, we are told that no two of the remaining 4 boxes should contain an equal number of balls. So, they should contain 2, 3, 4 and 5 balls each (also minimum possible).
Total = 1 + 1 + 1 + 2 + 3 + 4 + 5 = 17
Required number of balls = 17
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Question for Practice Questions Level 3: Permutation & Combination - 1
Try yourself:In how many ways can 9 blocks of 3 different colours and 3 different sizes, where all blocks of a particular colour have different sizes, be arranged in a straight line such that blocks of the same colour sit together, and at the point where the colour changes, the adjoining blocks are the same size?
Explanation
Let the colours be green (G), white (W) and blue (B) and the sizes be 1, 2 and 3. So, we have to arrange the 9 blocks: G1, G2, G3, W1, W2, W3, B1, B2, B3.First, the colours G, W and B can be arranged in 3! ways.
Let one such arrangement be G, W, B. Among green blocks, we have to arrange G1, G2 and G3.
We can select one size to be placed next to the first white block.
This can be done in 3C1 ways and the remaining sizes can be arranged in 2 ways.
Let one such arrangement be G1, G2, G3.
Then, W3 must come in fourth position, and we have to arrange W1 and W2 in fifth and sixth positions in any order.
This can be done in 2 ways.
Let the arrangement obtained so far be G1, G2, G3, W3, W1, W2.
Then, B2 must come in seventh position, and B1 and B3 can be arranged in 2 ways.
Total number of ways = 3! x 3C1 x 2 x 2 x 2 = 144
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