Combinations?

Question

From group of 6 girls and 7 boys, hoy many 5-member communities consist of more boys than girls?
Please try the explain the working also.....

Answer 1

In this problem we have a few combinations to calculate. A 5-member community with more boys that girls can have the possible scenarios:

5 boys, 0 girls
4 boys, 1 girl
3 boys, 2 girls

Now we need to calculate the combinations of this problem. A combination (n choose k) is defined by the following:

[n choose k] = n!/((n-k)!k!) where:

n = # of possible chocies
k = # of successes

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For 5 boys, 0 girls the following combination is:

[7 choose 5], because we have 7 boys to choose from out of 5 possible slots. This becomes:

[7 choose 5] = 7!/(2!5!) = 7*6/2 = 21 ways

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For 4 boys, 1 girl, we need to calculate 2 separate combinations. For the boys, we have [7 choose 4] and for the girls we have [6 choose 1]

[7 choose 4] = 7!/(4!3!) = 7*6*5/6 = 35 ways
[6 choose 1] = 6!/(5!1!) = 6 ways

Since the selection of boys and girls are indpendent of each other, we can multiply the separate combinations to get the total combinations.

4 boys, 1 girl = 35*6 = 210 ways

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The same reasoning can be made for 3 boys, 2 girls. For the boys, we have [7 choose 3] and the girls, we have [6 choose 2]

[7 choose 3] = 7!/(4!3!) = 7*6*5/6 = 35 ways
[6 choose 2] = 6!/(4!2!) = 6*5/2 = 15 ways

Thus, 3 boys, 2 girls = 35*15 = 525 ways

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Therefore, the total # of possible cominations is: 21+210+525 =

756 possible ways

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Hope this helps

Answer 2

We can have either 5 boys(0 girls), 4 boys(1 girl), or 3 boys(2 girls). So we have 7C5 + (7C4)*(6C1)+(7C3)*(6C2) = 21 + 210 + 525 = 756

Steve