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Permutations and Combinations: Mathematics | JEE Main

Suppose Anil's mother wants to give 5 whole fruits to Anil from a basket of 7 red apples, 5 white apples and 8 oranges. If in the selected 5 fruits, at least 2 oranges, at least one red apple and at least one white apple must be given, then the number of ways, Anil's mother can offer 5 fruits to Anil is

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Correct answer

The total number of ways to select 5 fruits satisfying all minimum conditions is 6860.

Option analysis

Why each option works or fails

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Solution

StepWorking

01given

A basket contains $7$ distinct red apples ( $R$ ), $5$ distinct white apples ( $W$ ), and $8$ distinct oranges ( $O$ ). Total fruits to select $= 5$ , with constraints: number of oranges $n_O \ge 2$ , red apples $n_R \ge 1$ , and white apples $n_W \ge 1$ .

02goal

Find the total number of ways to choose $5$ fruits such that $n_R + n_W + n_O = 5$ under the conditions $n_R \ge 1$ , $n_W \ge 1$ , and $n_O \ge 2$ .

03approach

Since $n_R + n_W + n_O = 5$ with $n_R \ge 1$ , $n_W \ge 1$ , $n_O \ge 2$ , the minimum total is $1 + 1 + 2 = 4$ . There is only $1$ fruit left to distribute. This yields exactly three mutually exclusive cases for the triplet $(n_R, n_W, n_O)$ : Case 1: $(1, 1, 3)$ Case 2: $(2, 1, 2)$ Case 3: $(1, 2, 2)$ .

04execute

Compute each case using combinations from distinct fruits: Case 1: $\binom{7}{1} \times \binom{5}{1} \times \binom{8}{3} = 7 \times 5 \times 56 = 1960$ Case 2: $\binom{7}{2} \times \binom{5}{1} \times \binom{8}{2} = 21 \times 5 \times 28 = 2940$ Case 3: $\binom{7}{1} \times \binom{5}{2} \times \binom{8}{2} = 7 \times 10 \times 28 = 1960$ Total ways $= 1960 + 2940 + 1960 = 6860$ .

✓verify

Check symmetry between Case 1 and Case 3, or group terms: $\binom{8}{2} [\binom{7}{2}\binom{5}{1} + \binom{7}{1}\binom{5}{2}] + \binom{8}{3}\binom{7}{1}\binom{5}{1} = 28 \times [105 + 70] + 56 \times 35 = 28 \times 175 + 1960 = 4900 + 1960 = 6860$ .

Hints that build this answer step by step

Which distributions of (Oranges, Red Apples, White Apples) satisfy the condition of selecting 5 fruits with at least 2 oranges, at least 1 red apple, and at least 1 white apple?
(2, 2, 1), (2, 1, 2), and (3, 1, 1)
How many ways can we select fruits for the cases where exactly 2 oranges are chosen: (2, 2, 1) and (2, 1, 2)?
4900 ways
What is the total number of ways to choose the 5 fruits across all three valid distributions?
6860

✓ 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

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Students also ask

Are fruits of the same type treated as distinct or identical?

In standard Indian entrance exams (JEE), real-world objects like fruits are treated as distinct unless explicitly stated as identical. That is why combinations $\binom{n}{r}$ are used.

Answer

The total number of ways to select 5 fruits satisfying all minimum conditions is 6860.

Why each option works or fails

Step-by-step solution

given: A basket contains $7$ distinct red apples ( $R$ ), $5$ distinct white apples ( $W$ ), and $8$ distinct oranges ( $O$ ). Total fruits to select $= 5$ , with constraints: number of oranges $n_O \ge 2$ , red apples $n_R \ge 1$ , and white apples $n_W \ge 1$ .
goal: Find the total number of ways to choose $5$ fruits such that $n_R + n_W + n_O = 5$ under the conditions $n_R \ge 1$ , $n_W \ge 1$ , and $n_O \ge 2$ .
approach: Since $n_R + n_W + n_O = 5$ with $n_R \ge 1$ , $n_W \ge 1$ , $n_O \ge 2$ , the minimum total is $1 + 1 + 2 = 4$ . There is only $1$ fruit left to distribute. This yields exactly three mutually exclusive cases for the triplet $(n_R, n_W, n_O)$ : Case 1: $(1, 1, 3)$ Case 2: $(2, 1, 2)$ Case 3: $(1, 2, 2)$ .
execute: Compute each case using combinations from distinct fruits: Case 1: $\binom{7}{1} \times \binom{5}{1} \times \binom{8}{3} = 7 \times 5 \times 56 = 1960$ Case 2: $\binom{7}{2} \times \binom{5}{1} \times \binom{8}{2} = 21 \times 5 \times 28 = 2940$ Case 3: $\binom{7}{1} \times \binom{5}{2} \times \binom{8}{2} = 7 \times 10 \times 28 = 1960$ Total ways $= 1960 + 2940 + 1960 = 6860$ .
verify: Check symmetry between Case 1 and Case 3, or group terms: $\binom{8}{2} [\binom{7}{2}\binom{5}{1} + \binom{7}{1}\binom{5}{2}] + \binom{8}{3}\binom{7}{1}\binom{5}{1} = 28 \times [105 + 70] + 56 \times 35 = 28 \times 175 + 1960 = 4900 + 1960 = 6860$ .

Permutations and Combinations: Mathematics | JEE Main

What feels right?

The total number of ways to select 5 fruits satisfying all minimum conditions is 6860.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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