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perm and comb question (1 Viewer)

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This topic really bugs me... so any help with explanation / hints with this question will be greatly appreciated.

" In how many ways can 8 boys be divided into two sets containing 5 and 3 respectively."

I just don't understand how to approach questions like this, and I got a feeling it's an easy combinations question as it's only Q6 in Fitzpatrick

Thanks in advance :)


Ahhhh another one that I've been staring at >.<

"A committee of 6 is selected from a 10 people of whom A and B are two... How many committees can be formed excluding A if B is included"
 
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Carrotsticks

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For the first one:

We have 8 boys so let's choose 3 to join a group. Hence, we have 8C3. Since the other 5 must go into the other group, we have 5C5. Hence the answer is 8C3 * 5C5.

For the second one:

If A is excluded, then we actually have a committee of 6 selected from a total of 9 people. However, since B is included, then we actually have a committee of 5 people selected from a total of 8 people, since B is guaranteed a spot. Hence, the answer is 8C5.
 

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