Questions and Answers Type: | MCQ (Multiple Choice Questions). |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Permutation and Combination Aptitude Questions and Answers. |

Number of Questions: | 10 Questions with Solutions. |

- How many integers between 100 and 1000 have no digits other than 3, 4, and 5?
- 24
- 27
- 36
- 42

Answer: (b) 27

Solution: Any number between 100 and 1000 is of 3 digits. So, we can fill the unit place, tenth place, and hundred's place by all the three digits 3, 4, and 5, means each place by 3 ways.

Hence, the number of integers, $$ = 3 \times 3 \times 3 $$ $$ = 27 $$

Solution: Any number between 100 and 1000 is of 3 digits. So, we can fill the unit place, tenth place, and hundred's place by all the three digits 3, 4, and 5, means each place by 3 ways.

Hence, the number of integers, $$ = 3 \times 3 \times 3 $$ $$ = 27 $$

- How many numbers are there between 1000 and 10000 in which all the digits are distinct?
- 4536
- 4332
- 4320
- 4263

Answer: (a) 4536

Solution: Any number between 1000 and 10000 is of four digits. Here, the numbers must have distinct digits. Here, repetition of digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is not allowed and zero (0) cannot be filled on the extream left end.

Now, thousand's place can be filled by 9 ways.

hundredth place can be filled by 9 ways.

tenth place can be filled by 8 ways.

unit place can be filled by 7 ways.

Hence total four digit numbers,$$ = 9 \times 9 \times 8 \times 7 $$ $$ = 4536 $$

Solution: Any number between 1000 and 10000 is of four digits. Here, the numbers must have distinct digits. Here, repetition of digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is not allowed and zero (0) cannot be filled on the extream left end.

Now, thousand's place can be filled by 9 ways.

hundredth place can be filled by 9 ways.

tenth place can be filled by 8 ways.

unit place can be filled by 7 ways.

Hence total four digit numbers,$$ = 9 \times 9 \times 8 \times 7 $$ $$ = 4536 $$

- How many ways 5 candidates can be filled for 2 posts?
- 10
- 20
- 30
- 40

Answer: (b) 20

Solution: The first post can be filled in 5 ways, the second post can be filled in 4 ways.

Now, the total number of ways posts can be filled, $$ = 5 \times 4 $$ $$ = 20 \ Ways $$

Solution: The first post can be filled in 5 ways, the second post can be filled in 4 ways.

Now, the total number of ways posts can be filled, $$ = 5 \times 4 $$ $$ = 20 \ Ways $$

- How many four digit odd numbers can be formed from the digits 1, 2, 3, 4, 5, 6, and 7 if the repetition of digits is not allowed?
- 220
- 360
- 480
- 550

Answer: (c) 480

Solution: Like the repetition of the digits is not allowed, and we have to form four digit odd numbers, so it is clear, that the unit place digit must be odd. So, we can fill the unit place by 1, 3, 5, and 7, that is in 4 ways.

Now, tenth digit can be filled by any of the remaining 6 numbers.

hundredth place can be filled by any of the remaining 5 numbers.

thousand's place can be filled by the remaining 4 numbers.

Hence, total number of ways,$$ = 4 \times 6 \times 5 \times 4 $$ $$ = 480 \ Ways $$

Solution: Like the repetition of the digits is not allowed, and we have to form four digit odd numbers, so it is clear, that the unit place digit must be odd. So, we can fill the unit place by 1, 3, 5, and 7, that is in 4 ways.

Now, tenth digit can be filled by any of the remaining 6 numbers.

hundredth place can be filled by any of the remaining 5 numbers.

thousand's place can be filled by the remaining 4 numbers.

Hence, total number of ways,$$ = 4 \times 6 \times 5 \times 4 $$ $$ = 480 \ Ways $$

- How many ways 10 students can be filled for three seats in a school?
- 525
- 635
- 680
- 720

Answer: (d) 720

Solution: The first seat can be filled by 10 ways, second place can be filled by 9 ways, and third place can be filled by 8 ways.

Now, total number of ways, seats can be filled,$$ = 10 \times 9 \times 8 $$ $$ = 720 \ Ways $$

Solution: The first seat can be filled by 10 ways, second place can be filled by 9 ways, and third place can be filled by 8 ways.

Now, total number of ways, seats can be filled,$$ = 10 \times 9 \times 8 $$ $$ = 720 \ Ways $$

- In how many ways can 3 sportsmen be selected from a group of 8?
- 56
- 62
- 68
- 72

Answer: (a) 56

Solution: 3 sportsmen can be selected from the group of 8 by \(8C_3\) ways.

Hence, $$ 8C_3 = \frac{8!}{(8 - 3)! \times 3!} $$ $$ = \frac{8!}{5! \times 3!} $$ $$ = 56 \ Ways $$

Solution: 3 sportsmen can be selected from the group of 8 by \(8C_3\) ways.

Hence, $$ 8C_3 = \frac{8!}{(8 - 3)! \times 3!} $$ $$ = \frac{8!}{5! \times 3!} $$ $$ = 56 \ Ways $$

- How many three digit odd numbers can be formed from the digits 1, 2, 3, 4, and 5 if the repetition of digits is allowed?
- 36
- 42
- 48
- 56

Answer: (c) 48

Solution: Here, the repetition of the digits is allowed and we have to form three digit odd numbers, so it is clear that the unit place can be filled by 1, 3, and 5, that is in 3 ways.

Now, if the repetition of digits is allowed, so we can fill the tenth and hundredth place by any of the remaining four digits in 4 ways each.

Hence, $$ = 4 \times 4 \times 3 $$ $$ = 48 \ Ways $$

Solution: Here, the repetition of the digits is allowed and we have to form three digit odd numbers, so it is clear that the unit place can be filled by 1, 3, and 5, that is in 3 ways.

Now, if the repetition of digits is allowed, so we can fill the tenth and hundredth place by any of the remaining four digits in 4 ways each.

Hence, $$ = 4 \times 4 \times 3 $$ $$ = 48 \ Ways $$

- How many ways 5 different colour of flowers can be arranged in a garland?
- 12
- 15
- 16
- 18

Answer: (a) 12

Solution: It can be arranged by, $$ = \frac{(5 - 1)!}{2} $$ $$ = \frac{4!}{2} $$ $$ = 12 \ Ways $$

Solution: It can be arranged by, $$ = \frac{(5 - 1)!}{2} $$ $$ = \frac{4!}{2} $$ $$ = 12 \ Ways $$

- How many different garlands can be formed with 4 Red and 5 Yellow flowers?
- 4
- 5
- 6
- 7

Answer: (d) 7

Solution: Total number of flowers,$$ = 4 + 5 $$ $$ = 9 $$ Hence, total number of garlands, $$ = \frac{(9 - 1)!}{2} \times \frac{1}{4! \times 5!} $$ $$ = \frac{8!}{2 \times 4! \times 5!} $$ $$ = 7 \ Garlands $$

Solution: Total number of flowers,$$ = 4 + 5 $$ $$ = 9 $$ Hence, total number of garlands, $$ = \frac{(9 - 1)!}{2} \times \frac{1}{4! \times 5!} $$ $$ = \frac{8!}{2 \times 4! \times 5!} $$ $$ = 7 \ Garlands $$

- How many ways 10 different colour and design of beads can be arranged in a neckless?
- 185,256
- 182,335
- 181,440
- 179,120

Answer: (c) 181,440

Solution: It can be arranged by,$$ = \frac{(10 - 1)!}{2} $$ $$ = \frac{9!}{2} $$ $$ = 181,440 \ Ways $$

Solution: It can be arranged by,$$ = \frac{(10 - 1)!}{2} $$ $$ = \frac{9!}{2} $$ $$ = 181,440 \ Ways $$

Lec 1: Permutation and Combination
Lec 2: Binomial Theorem
Exercise-1
Exercise-2
Exercise-3
Exercise-4
Exercise-5