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Question

Difficulty: MediumCombinations

In a mathematics examination paper consisting of 88 questions, a candidate is required to answer 55 questions in total. If the candidate must answer the first 22 questions, in how many ways can the candidate select the remaining questions?

Answer: 20 ways

Answer

The candidate can select the questions in 20 ways.
Because the first 2 questions are mandatory, the choice is reduced to picking 3 additional questions from the remaining 6 questions. Since selection order is irrelevant, the number of distinct ways is \(\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20\).

Step-by-Step Solution

1
Determine the remaining number of questions needed
3 questions
Since 2 out of the required 5 questions are compulsory, the candidate must choose 3 more.
2
Determine the available pool of remaining questions
6 questions
Subtracting the 2 compulsory questions from the 8 total questions leaves 6 questions available.
3
Calculate the combinations using the nCr formula
20 ways
The order in which questions are selected does not matter, so we use combinations: \(\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20\).

Key Concept

Combinations with restricted or fixed choices
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Topics

Combinations

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