Question 1

In how many ways can 3 letters be placed in 3 addressed envelopes so that no letter goes into its correct envelope?

Accepted Answer

2. This is the derangement of 3 objects, D₃ = 2 (for letters a, b, c the only ways are bca and cab).

Question 2

In how many ways can 4 letters be placed in 4 addressed envelopes so that every letter goes into a wrong envelope?

Accepted Answer

9. The number of derangements of 4 objects is D₄ = 9.

Question 3

How many 4-digit numbers have their digits in strictly increasing order (e.g. 1359)?

Accepted Answer

126. Choose any 4 digits from 1–9; they can be arranged in increasing order in exactly one way: ⁹C₄ = 126.

Question 4

How many 3-digit numbers divisible by 5 can be formed using the digits 1, 2, 3, 4, 5 without repetition?

Accepted Answer

12. The units digit must be 5; the first two places are filled from the remaining 4 digits: 4 × 3 = 12.

Question 5

How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6, 7 without repetition?

Accepted Answer

90. Units digit is 2, 4 or 6 (3 ways); the other two places take 6 × 5 = 30 ways. Total 3 × 30 = 90.

Question 6

The sum of all 3-digit numbers formed by using the digits 1, 2, 3 exactly once each is:

Accepted Answer

1332. Each digit occupies each place 2! = 2 times, so the sum is (1 + 2 + 3) × 2 × 111 = 1332.

Question 7

In how many ways can 5 people be seated in a row if two particular persons refuse to sit next to each other?

Accepted Answer

72. Total arrangements minus together: 5! − 2! × 4! = 120 − 48 = 72.

Question 8

In how many ways can 4 books be selected from 10 different books if a particular book is always included?

Accepted Answer

84. With that book fixed, choose 3 more from the remaining 9: ⁹C₃ = 84.

Question 9

In how many ways can 3 players be selected from 11 players if one particular player is never selected?

Accepted Answer

120. Exclude that player and choose 3 from the remaining 10: ¹⁰C₃ = 120.

Question 10

How many numbers greater than 3000 can be formed using all the digits 1, 2, 3, 4 exactly once each?

Accepted Answer

12. The number is 4-digit; the leading digit must be 3 or 4 (2 ways), and the rest arrange in 3! = 6 ways: 2 × 6 = 12.

Question 11

How many 4-digit numbers divisible by 10 can be formed from the digits 0, 1, 2, 3, 4, 5 without repetition?

Accepted Answer

60. The units digit must be 0; the other three places are filled from the remaining 5 digits: 5 × 4 × 3 = 60.

Question 12

From 8 students, 3 are to be chosen and appointed as president, vice-president and secretary (one post each). The number of ways is:

Accepted Answer

336. Choose 3 and then assign the distinct posts: ⁸C₃ × 3! = 56 × 6 = 336 (which equals ⁸P₃).

Question 13

If all permutations of the letters of the word RANK are arranged in dictionary order, the rank (position) of the word RANK is:

Accepted Answer

20. Words starting with A, K, N come first: 3 × 3! = 18. Then among R-words, RAK_ (1 word) precedes RANK, so 18 + 1 = 19 words come before it and its rank is 20.

Question 14

If all permutations of the letters of the word MOTHER are arranged as in a dictionary, the rank of the word MOTHER is:

Accepted Answer

309. Counting smaller words position by position: 2×120 + 2×24 + 3×6 + 1×2 + 0 + 0 = 308 words precede it, so the rank is 309.

Question 15

All permutations of the letters A, B, C, D, E are written and arranged in dictionary order. The 60th word in the list is:

Accepted Answer

CBEDA. Words 1–24 start with A, 25–48 with B; the 60th is the 12th C-word. Among C-words, the 7th–12th start CB, and the 12th is the last of them: CBEDA.

Question 16

How many numbers greater than 23000 can be formed using all the digits 1, 2, 3, 4, 5 exactly once each?

Accepted Answer

90. Leading digit 3, 4 or 5: 3 × 4! = 72. Leading digit 2 needs second digit 3, 4 or 5: 3 × 3! = 18. Total 72 + 18 = 90.

Question 17

The sum of all 4-digit numbers formed by using the digits 1, 2, 3, 4 exactly once each is:

Accepted Answer

66660. Each digit occupies each place 3! = 6 times: sum = (1 + 2 + 3 + 4) × 6 × 1111 = 66660.

Question 18

How many 4-digit even numbers can be formed using the digits 0, 1, 2, 3, 4 without repetition?

Accepted Answer

60. Units 0: 4 × 3 × 2 = 24. Units 2 or 4: leading digit has 3 choices (not 0), middle places 3 × 2, giving 2 × 3 × 6 = 36. Total 24 + 36 = 60.

Question 19

From 8 friends, 5 are to be invited to dinner, but two particular friends must be either both invited or both left out. The number of ways is:

Accepted Answer

26. Both in: choose 3 from the other 6 = ⁶C₃ = 20. Both out: choose 5 from 6 = ⁶C₅ = 6. Total 20 + 6 = 26.

Question 20

From 5 vowels and 6 consonants, how many 5-letter words (with or without meaning) can be formed using exactly 2 vowels and 3 consonants?

Accepted Answer

24000. Select then arrange: ⁵C₂ × ⁶C₃ × 5! = 10 × 20 × 120 = 24000.

Miscellaneous Applications & Advanced Word Problems — JEE Main Mathematics MCQs with Solutions

Questions with solutions