# Of the 20 members of a cricket club 2 are wicket keepers and 5 bowlers. We have to select a team of 11 from this lot if (a) the team is to include only 1 wicket keeper and at least 3 bowler? (b) if 2 wicket keepers, 5 and 3 bowlers are included?

## Answer is - 54054 Plz show how to do it

See below:

#### Explanation:

We have a cricket club with 20 members: 2 wicket keepers, 5 bowlers, and 13 general members.

With picking a team, we're looking at combination calculations:

C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

a

We want 1 wicket keeper and at least 3 bowlers.

To do this, let's first look at the wicket keeper. There are 2 of them and we want 1, which gives:

$\left(\begin{matrix}2 \\ 1\end{matrix}\right) = 2$

We want at least 3 bowlers. Let's pick 3 and then if more show up in the final pick, that's fine:

$\left(\begin{matrix}5 \\ 3\end{matrix}\right) = 10$

And now the rest of the team. We don't want the other wicket keeper, so he's excluded. We've also already picked 4 people so they're excluded. We need 7 people to round out the team and there are 15 members left we can pick from $\left(20 - 1 - 4\right)$:

$\left(\begin{matrix}15 \\ 7\end{matrix}\right) = 6435$

And so we have:

$2 \times 10 \times 6435 = \text{128,700}$

b

This part of the question isn't clear, so I'll answer the question as if it's asking for exactly 3 of the 5 bowlers are picked.

We first need to pick both wicket keepers:

$\left(\begin{matrix}2 \\ 2\end{matrix}\right) = 1$

And exactly 3 bowlers:

$\left(\begin{matrix}5 \\ 3\end{matrix}\right) = 10$

What changes, however, is that now we don't want to pick the final 7 players from 15 possible players (as we did in a) because we could end up with more bowlers than asked for. So instead we pick from the 13 general members:

$\left(\begin{matrix}13 \\ 7\end{matrix}\right) = 1716$

Giving:

$1 \times 10 \times 1716 = \text{17,160}$