# Find out in how many ways the operation can be performed?

i) In how many ways can a committee of $5$ or more be formed from 12 persons?

ii) In how many ways can a committee of $5$ be formed from 12 persons if only two of a group of $3$ persons must always be included?

All I know is it must be combination cause order/arrangement does not matter.

I know to solve by formula too; but confused how to begin. Help...

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To make a comittee of $k$ out of a total of $n$ is selecting a set of $k$ out of the $n$, which can be done in $\binom{n}{k} = \frac{n!}{k! (n - k)!}$ ways. To see why, you first select the $k$ members in order. The first one you can select in $n$ ways, the second in $n - 1$ ways (the one selected first isn't eligible anymore), ..., the $k$-th in $n - k + 1$ ways. In all, this gives $n (n - 1) \ldots (n - k + 1) = \frac{n!}{(n - k)!}$ ways. But for the comittee the order is irrelevant, you overcounted by a factor of $k!$ by ordering the people. So the number of $k$-subsets of an $n$-set is as stated above.
i) In how many ways can a committee of $5$ or more be formed from $12$ persons?
## Ans: $^{12}C_{5} + ^{12}C_6 + ^{12}C_7 + ... + ^{12}C_{12}$
ii) In how many ways can a committee of 5 be formed from $12$ persons if only two of a group of $3$ persons must always be included? Ans: $^{3}C_2*^{9}C_3 = 3*84 = 252$