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4 years ago

7

Given a group of 8 women and 11 men, how many different ways are there of choosing one man and one woman for a committee?

1 answer:

Mashutka [201]4 years ago

8 0

Any of the 11 man can be chosen, and combined with any of the 8 women.

Assume we select man1. The selected committee can be:

(m1,w1), (m1,w2), (m1,w3), (m1,w4), (m1,w5), (m1,w6), (m1,w7), (m1,w8),

so there are 8 committees selections with man1 in them.

we could repeat the same procedure for the remaining 10 men, and get 8 committees where each of them is a member.

so there are 11*8=88 ways of choosing 1 man and 1 woman.

Answer: 88

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<em>See the image in the attachment for the referred diagram.</em>

<em />

The two triangles, triangle AEC and triangle BDC are similar triangles.
Therefore, the ratio of the corresponding sides of triangles AEC and BDC will be the same.

<em>This implies that</em>:

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<em><u>Given:</u></em>

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Cross multiply

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Thus:

$AB = AC - BC$

Substitute

$AB = 6.15 - 4.1\\\\AB = 2.05 $ cm$

Therefore, by applying the knowledge of similar triangles, the lengths of AE and AB are:

a. $\mathbf{AE = 3.9 $ cm}\\\\$

b. $\mathbf{AB = 2.05 $ cm} \\\\$

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