There are 3 wicketkeepers and 5 bowlers among 22 cricket players. A team of 11 players is to be selected so that there is exactly one wicketkeeper and at least 4 bowlers in the team.

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Concept understanding — Combination

Combination: Choosing Without Ordering

Imagine you have three friends: A, B, and C. You need to pick two of them to form a committee. How many different committees can you make?

You could pick A and B, or A and C, or B and C. That's three committees. Notice something crucial: the committee {A, B} is exactly the same as {B, A}. Who gets picked first doesn't matter — only who is on the committee matters. This is the core idea of a combination: a selection where the order of selection does not matter.

The Intuition: "How many groups?"

A combination answers the question: "From a set of $n$ distinct items, how many ways can I choose $r$ of them, ignoring the order in which I pick them?"

Think of it as forming a team, a committee, a hand of cards, or a set of ingredients for a recipe. In all these cases, swapping two members around doesn't give you a new team — it's still the same group.

The Precise Statement

A combination is a selection of $r$ objects from a set of $n$ distinct objects, where the order of selection does not matter. The number of such combinations is denoted by:

\binom{n}{r} \quad \text{or} \quad ^nC_r

and is given by the formula:

\binom{n}{r} = \frac{n!}{r! \, (n-r)!}

where $n!$ (read "n factorial") means $n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$ , and $0! = 1$ by definition.

Why the $r!$ in the denominator?

This is the key to understanding combinations. If order did matter, the number of ways to pick $r$ objects from $n$ would be $n \times (n-1) \times \cdots \times (n-r+1)$ , which is written as $\frac{n!}{(n-r)!}$ . That's called a permutation.

But in a combination, every group of $r$ objects can be rearranged in $r!$ different orders. Since we don't care about order, we divide by $r!$ to "collapse" all those orderings into one single group.

So:

\binom{n}{r} = \frac{\text{number of ordered selections}}{\text{number of ways to order each selection}} = \frac{n!}{(n-r)!} \div r! = \frac{n!}{r! \, (n-r)!}

A Quick Example

From a class of 10 students, how many ways can you choose a team of 3?

Here $n=10$ , $r=3$ .

\binom{10}{3} = \frac{10!}{3! \, 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120

So there are 120 different teams possible. …

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Of the 22 players, 3 are wicketkeepers, 5 are bowlers, leaving $22-3-5=14$ 'other' players (batsmen/all-rounders). Exactly 1 wicketkeeper: ${}^3C_1=3$ ways. Bowlers can be 4 or 5 (at least 4 out of the 5 available). Bowlers=4: remaining team slots $=11-1-4=6$ from the 14 others: ${}^5C_4\times{}^{14}C_6=5\times3003=15015$ ; combined with the 3 wicketkeeper choices: $3\times15015=45045$ . Bowlers=5: remaining slots …

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Q.There are 3 wicketkeepers and 5 bowlers among 22 cricket players. A team of 11 players is to be selected so that there is exactly one wicketkeeper and at least 4 bowlers in the team.

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