JEE MainMathematics

Reviewed by official_key

+4 marks−1 if incorrectNumericalpyq

Permutations and Combinations: Mathematics | JEE Main

A boy needs to select five courses from

12

available courses, out of which

5

courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Type the value - units or words beside it are fine.

Step-by-step solutionView

Correct answer

The number of ways to choose five courses such that at most two are language courses is 546.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Total courses = $12$ , of which $5$ are language courses and $12 - 5 = 7$ are non-language courses. We need to select $5$ courses in total with at most $2$ language courses.

02goal

Find the total number of ways to choose $5$ courses satisfying the condition of selecting at most $2$ language courses.

03approach

Split the requirement 'at most 2 language courses' into three mutually exclusive cases: 0 language courses, 1 language course, and 2 language courses. For each case, choose the remaining required courses from the 7 non-language courses and sum the combinations.

04execute

Case 1: $0$ language and $5$ others $\implies \binom{5}{0} \binom{7}{5} = 1 \times 21 = 21$ . Case 2: $1$ language and $4$ others $\implies \binom{5}{1} \binom{7}{4} = 5 \times 35 = 175$ . Case 3: $2$ languages and $3$ others $\implies \binom{5}{2} \binom{7}{3} = 10 \times 35 = 350$ . Total ways $= 21 + 175 + 350 = 546$ .

✓verify

Using complementary counting: Total ways without restrictions $= \binom{12}{5} = 792$ . Unfavorable cases: $3$ language courses $\implies \binom{5}{3}\binom{7}{2} = 10 \times 21 = 210$ ; $4$ language courses $\implies \binom{5}{4}\binom{7}{1} = 5 \times 7 = 35$ ; $5$ language courses $\implies \binom{5}{5}\binom{7}{0} = 1 \times 1 = 1$ . Total unfavorable $= 210 + 35 + 1 = 246$ . Desired ways $= 792 - 246 = 546$ , which matches.

Hints that build this answer step by step

How can the condition of selecting 'at most two language courses' out of five total courses be partitioned into mutually exclusive cases?
0 language courses, 1 language course, or 2 language courses
Given 5 language courses and 7 non-language courses, which sum of combinations represents the three valid cases?
C(5,0)*C(7,5) + C(5,1)*C(7,4) + C(5,2)*C(7,3)
What are the numerical values of the three terms C(5,0)*C(7,5), C(5,1)*C(7,4), and C(5,2)*C(7,3)?
21, 175, and 350
What is the total number of ways to choose the five courses?
546

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we multiply within each case and add across cases?

Within a case, choosing languages and non-languages are independent sequential choices (multiplication principle). The three cases are mutually exclusive alternatives (addition principle).

Answer

The number of ways to choose five courses such that at most two are language courses is 546.

Why each option works or fails

Step-by-step solution

given: Total courses = $12$ , of which $5$ are language courses and $12 - 5 = 7$ are non-language courses. We need to select $5$ courses in total with at most $2$ language courses.
goal: Find the total number of ways to choose $5$ courses satisfying the condition of selecting at most $2$ language courses.
approach: Split the requirement 'at most 2 language courses' into three mutually exclusive cases: 0 language courses, 1 language course, and 2 language courses. For each case, choose the remaining required courses from the 7 non-language courses and sum the combinations.
execute: Case 1: $0$ language and $5$ others $\implies \binom{5}{0} \binom{7}{5} = 1 \times 21 = 21$ . Case 2: $1$ language and $4$ others $\implies \binom{5}{1} \binom{7}{4} = 5 \times 35 = 175$ . Case 3: $2$ languages and $3$ others $\implies \binom{5}{2} \binom{7}{3} = 10 \times 35 = 350$ . Total ways $= 21 + 175 + 350 = 546$ .
verify: Using complementary counting: Total ways without restrictions $= \binom{12}{5} = 792$ . Unfavorable cases: $3$ language courses $\implies \binom{5}{3}\binom{7}{2} = 10 \times 21 = 210$ ; $4$ language courses $\implies \binom{5}{4}\binom{7}{1} = 5 \times 7 = 35$ ; $5$ language courses $\implies \binom{5}{5}\binom{7}{0} = 1 \times 1 = 1$ . Total unfavorable $= 210 + 35 + 1 = 246$ . Desired ways $= 792 - 246 = 546$ , which matches.

Permutations and Combinations: Mathematics | JEE Main

What feels right?

The number of ways to choose five courses such that at most two are language courses is 546.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

We think you should solve this next

Eight persons are to be transported from city A to city B in three cars of different make…

Number of 4-digit numbers that are less than or equal to $2800$ and either divisible by…

If ${ }^{2 n+1} \mathrm{P}_{n-1}:{ }^{2 n-1} \mathrm{P}_n=11: 21$ , then $n^2+n+15$ is equ…

Students also ask

Permutations and Combinations: Mathematics | JEE Main

The number of ways to choose five courses such that at most two are language courses is 546.

01given

02goal

03approach

04execute

✓verify

Eight persons are to be transported from city A to city B in three cars of different make…

Number of 4-digit numbers that are less than or equal to 280028002800 and either divisible by…

If 2n+1Pn−1:2n−1Pn=11:21{ }^{2 n+1} \mathrm{P}_{n-1}:{ }^{2 n-1} \mathrm{P}_n=11: 212n+1Pn−1​:2n−1Pn​=11:21, then n2+n+15n^2+n+15n2+n+15 is equ…

Students also ask

Number of 4-digit numbers that are less than or equal to $2800$ and either divisible by…

If ${ }^{2 n+1} \mathrm{P}_{n-1}:{ }^{2 n-1} \mathrm{P}_n=11: 21$ , then $n^2+n+15$ is equ…