This set of Aptitude Questions and Answers (MCQs) focuses on “Permutations and Combinations – Set 5”.

1. A bag consists of 2 red balls, 3 green balls and 4 blue balls. In how many ways can 3 balls be drawn, if at least one of them is a blue ball?

a) 74

b) 76

c) 2! * 3! * 4!

d) 78

View Answer

Explanation: There are 2 red balls, 3 green balls and 4 blue balls.

The possible outcomes for at least one blue ball, when 3 balls are drawn from the bag are:

(1 red, 1 green, 1 blue), (1 red, 2 blue), (1 green, 2 blue), (2 red, 1 blue), (2 green, 1 blue), (3 blue).

Required number of ways = (

^{2}C

_{1}*

^{3}C

_{1}*

^{4}C

_{1}) + (

^{2}C

_{1}*

^{4}C

_{2}) + (

^{3}C

_{1}*

^{4}C

_{2}) + (

^{2}C

_{2}*

^{4}C

_{1}) + (

^{3}C

_{2}*

^{4}C

_{1}) + (

^{4}C

_{3}).

⟹ (2 * 3 * 4) + (2 * 6) + (3 * 6) + (1 * 4) + (3 * 4) + (4)

⟹ 24 + 12 + 18 + 4 + 12 + 4 = 74

Number of ways if at least one of them is blue = 74.

2. Three coins are tossed at a time. What is the probability of getting two heads?

a) 3 / 8

b) 1 / 2

c) 7 / 8

d) 2 / 3

View Answer

Explanation: When three coins are tossed, total number of outcomes will be as follows:

(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT) = 8.

We have to find the probability of getting at least one head.

Required type of outcomes = (HHH, HHT, HTH, THH).

Probability of getting at least one head = 4 / 8 = 1 / 2.

3. How many three digit numbers can be formed with the numbers 8, 2, 4, 5, 1 (if repetition of digits is not allowed)?

a) 3!

b) 5!

c) 60

d) 80

View Answer

Explanation: The numbers given are 8, 2, 4, 5, 1.

Here, repetition of digits is not allowed.

The first digit can be arranged in 5 ways.

Second digit can be arranged in 4 ways.

Third digit can be arranged in 3 ways.

Therefore, Number of three digit numbers can be formed = 5 * 4 * 3 = 60.

4. Find the number of new words that can be formed with the word ‘AMBITION’ all starting with A.

a) 8! / 2!

b) 8! * 2!

c) 7!

d) 7! / 2!

View Answer

Explanation: The word ‘AMBITION’ has 8 letters.

The word has 1’A’, 1’M’, 1’B’, 2I’s, 1’T’, 1’O’ and 1’N’.

We have 8 letters that can be arranged in 8! Ways.

The letter A is fixed as the starting letter. So, 7 letters should be arranged in 7! Ways.

The letter ‘I’ is repeated = 2!.

Number of words that can be formed with starting letter as ‘A’ = 7! / 2!.

5. How many numbers can be formed with numbers 1, 2, 4, 5, 8 without repetition?

a) 125

b) 5!

c) 325

d) 625

View Answer

Explanation: Number of one digit numbers we can form = 5.

Number of two digit numbers we can form = 5 * 4 = 20.

Number of three digit numbers we can form = 5 * 4 * 3 = 60.

Number of four digit numbers we can form = 5 * 4 * 3 * 2 = 120.

Number of five digit numbers we can form = 5 * 4 * 3 * 2 * 1 = 120.

Total number of numbers we can form = 5 + 20 + 60 + 120 + 120 = 325.

6. How many three digit numbers can be formed such that all the digits are even?

a) 5!

b) 125

c) 5! * 3!

d) 100

View Answer

Explanation: We have 5 even numbers, they are 0, 2, 4, 6, 8.

For the first digit we have 4 options 2, 4, 6, 8.

For the second digit we have 5 options 0, 2, 4, 6, 8.

For the third digit we have 5 options 0, 2, 4, 6, 8.

Hence, number of three digit numbers that can be formed with even numbers = 4 * 5 * 5 = 100.

7. From a pack of cards, half of the cards are selected. What will be the possibility of outcomes?

a) 52! / 26!

b) 1

c) ^{52}C_{25}

d) ^{52}C_{26}

View Answer

Explanation: Pack of cards has 52 cards in it.

Half of the cards are selected.

Half of 52 cards is 26.

Selecting 26 out of 52 =

^{52}C

_{26}.

The possible outcomes will be =

^{52}C

_{26}.

8. In how many ways can the word ‘ESPECIALLY’ be arranged such that the vowels always come together?

a) 42600

b) 25200

c) 10080

d) 30240

View Answer

Explanation: The word ‘ESPECIALLY’ has 10 letters.

We have to consider the vowels (EEIA) as 1 letter.

Now, we have SPCLLY and (EEIA) i.e., 7(6 + 1) letters.

In SPCLLY there are 1 S, 1 P, 1 C, 2L’s and 1 Y.

Number of ways we can arrange these letters = 7! / 2! = 2520.

There are 4 vowels, can be arranged = 4! / (2!)(1!)(1!) = 12.

Number of ways we can arrange so that vowels can come together = (2520 * 12) = 30240.

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