A committee consisting of 4 men and 4 women is to be selected from a group of 8 men and 6 women.
a) Find the number of different ways in which the committee can be selected.
b) Suppose that such a committee was selected. Find the number of different ways in which those committee members can be seated in a row if no two women can be seated next to each other
My Attempt –
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${8}\choose{4}$ $\times$ ${6}\choose{4}$ $ =1050$
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Since we have 4 men and 4 women,
$$4\times4\times3\times3\times2\times2 =576$$
Is there a different method to think the last part
Best Answer
While your first answer is correct, the second one is not. Here's one way to solve to second part.
We start with the number of ways in which we can seat the 4 men. That's 4!.
Now, when talking about the women, we have a restriction. No two of them can sit together. So, we address this condition as follow.
$$\times \, M \times M \times M \times M \,\times$$
What you see above is a representation of the 4 men sitting in a row (something we have taken care of earlier). Now, we can use the position indicated by $\times$ marks to seat the women. That way, there will always be at least one man between two women - as necessitated by the question.
So, we have 5 positions to seat 4 women. This can be dome in $^5P\,_4$ ways.
Hence, the total number of ways is
$$ \begin{align*} 4! \,\times \,^5P\,_4 &= 2880 \end{align*}$$