Combinations (1 Viewer)

moderntortoisecat · Feb 4, 2024

A committee of seven politicians is chosen from ten Liberal members, eight labor members and 5 independents. In how many ways can this be done to include exactly one independent, at least 3 liberal members and at least one labor member?

Average Boreduser · Feb 4, 2024

tabulate it and use multiplication rule

Luukas.2 · Feb 4, 2024

moderntortoisecat said:
A committee of seven politicians is chosen from ten Liberal members, eight labor members and 5 independents. In how many ways can this be done to include exactly one independent, at least 3 liberal members and at least one labor member?

$\bullet \ \text{1 Ind, 5 Libs, 1 Lab:} \quad \binom{5}{1} \binom{10}{5} \binom{8}{1}$

$\bullet \ \text{1 Ind, 4 Libs, 2 Lab:} \quad \binom{5}{1} \binom{10}{4} \binom{8}{2}$

$\bullet \ \text{1 Ind, 3 Libs, 3 Lab:} \quad \binom{5}{1} \binom{10}{3} \binom{8}{3}$

$\begin{align*} \text{Total combinations} &= \binom{5}{1} \binom{10}{5} \binom{8}{1} + \binom{5}{1} \binom{10}{4} \binom{8}{2} + \binom{5}{1} \binom{10}{3} \binom{8}{3} \\ &= \binom{5}{1} \left[\binom{10}{5} \binom{8}{1} + \binom{10}{4} \binom{8}{2} + \binom{10}{3} \binom{8}{3}\right] \\ &= 5 \left(252 \times 8 + 210 \times 28 + 120 \times 56\right) \\ &= 73\ 080 \end{align*}$

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Combinations (1 Viewer)

moderntortoisecat

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Average Boreduser

Rising Renewal

Luukas.2

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