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JEE • MATHEMATICS • PERMUTATION AND COMBINATION

Permutation and combination practice · Hard

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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1 Hard

1 of 7

Application +4 | -1

Number of words of 4 letters that can be formed with the letters of the word IIT JEE, is

View answer and explanation

Correct option: C

102

Option C (102) is correct because it accounts for all possible combinations of 4-letter words formed using the letters of the word IIT JEE, considering repetitions.

2 Hard

2 of 7

Analyzing +4 | -1

Let

p

be a prime number such that

p \geq 11

. Let

n=p !+1

. The number of primes in the list

n+1, n+2, n+3, \ldots n+p-1

, is

View answer and explanation

Correct option: D

None of these

The correct answer is D (None of these) because for each term in the list,

n+i

is divisible by

i+1

, making none of them prime.

3 Hard

3 of 7

Application +4 | -1

In an examination, the maximum marks for each of three papers is

n

and that for fourth paper is

2 n

. Then, the number of ways in which a candidate can get

3 n

marks, is

View answer and explanation

Correct option: B

\frac{1}{6}(n+1)\left(5 n^2+10 n+6\right)

Option B (

\frac{{1}}{{6}}(n+1)\left(5 n^2+10 n+6\right)

) is correct because it represents the number of ways a candidate can get 3n marks, considering the maximum marks for each paper.

4 Hard

4 of 7

Analyzing +4 | -1

\lim\limits_{n \to \infty } \frac{{\sum\limits_{1 \leqslant i} {\sum\limits_{ \leqslant j \leqslant n} {\left( {i + 1} \right)\left( {j + 1} \right)} } }}{{{n^4}}}

is equal to

View answer and explanation

Correct option: C

\frac{1}{8}

Option C (

\frac{1}{8}

) is the correct answer because the limit of the given expression as n approaches infinity is

\frac{1}{8}

5 Hard

5 of 7

Application +4 | -1

The number of ways of selecting

n

objects out of

3 n

objects, of which

n

are alike and rest are different, is

View answer and explanation

Correct option: A

2^{2 n-1}+\frac{(2 n-1) !}{n !(n-1) !}

Option A (

2^{2 n-1}+\frac{{(2 n-1) !}}{{n !(n-1) !}}

) is correct because it represents the number of ways to select n objects out of 3n objects, where n are alike and the rest are different.

6 Hard

6 of 7

Analyzing +4 | -1

Ten persons numbered

1,2,3, \ldots, 10

play a chess tournament, each player playing against every player exactly one game. It is known that no game ends in a draw. Let

w_1, w_2, w_3, \ldots, w_{10}

be the number of games won by player

1,2,3, \ldots, 10

respectively and

l_1, l_2, l_3, \ldots

l_{10}

be the number of games lost by the players

1,2,3, \ldots

, 10 respectively, then

View answer and explanation

Correct option: C

\Sigma w_i^2=\Sigma l_i^2

Option C (

\Sigma w_i^2=\Sigma l_i^2

) is correct because in a tournament where each game has a winner and a loser, the sum of the squares of the wins equals the sum of the squares of the losses.

7 Hard

7 of 7

Application +4 | -1

In how many ways 4 alphabets can be chosen from the word ASSASSINATION?

View answer and explanation

Correct option: B

Option D, 72, is correct because it represents the number of ways to choose 4 alphabets from the word ASSASSINATION.

Revision notes and key takeaways

Review PERMUTATION AND COMBINATION in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 7 single-correct-answer MCQs on PERMUTATION AND COMBINATION.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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