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A group consists of 5 men and 5 women. If the number of different five-person committees containing $\mathrm{k}$ men and $(5-\mathrm{k})$ women is 100, what is the value of $\mathrm{k}$ ?
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The correct answer is:
2 or 3
$\mathrm{K}$ men selected out of 5 and $5-\mathrm{k}$ women out of 5 . These are ${ }^{5} \mathrm{C}_{\mathrm{k}}$ and ${ }^{5} \mathrm{C}_{5-\mathrm{k}}$
According to problem:
${ }^{5} \mathrm{C}_{\mathrm{k}} \times{ }^{5} \mathrm{C}_{5-\mathrm{k}}=100$
$\Rightarrow \frac{5 !}{\mathrm{k} !(5-\mathrm{k}) !} \times \frac{5 !}{(5-\mathrm{k}) ! 5 !}=100$
$\Rightarrow\left(\frac{5}{k !(5-k) !}\right)^{2}=100$
$\Rightarrow \frac{5 !}{\mathrm{k} !(5-\mathrm{k}) !}=10$
This is true for $\mathrm{k}=2$ or 3 .
According to problem:
${ }^{5} \mathrm{C}_{\mathrm{k}} \times{ }^{5} \mathrm{C}_{5-\mathrm{k}}=100$
$\Rightarrow \frac{5 !}{\mathrm{k} !(5-\mathrm{k}) !} \times \frac{5 !}{(5-\mathrm{k}) ! 5 !}=100$
$\Rightarrow\left(\frac{5}{k !(5-k) !}\right)^{2}=100$
$\Rightarrow \frac{5 !}{\mathrm{k} !(5-\mathrm{k}) !}=10$
This is true for $\mathrm{k}=2$ or 3 .
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