Test Level 3: Permutation & Combination - 1 - CAT MCQ

(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.

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

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

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.

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

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!)

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

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)

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

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