Probability Q-combinatorics type

Sonal7

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I am finding a hard. I know that 24C4 will give you all the different combinations. Then i am lost as to what to do next.
 

pka

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#9 a) \(^{24}\mathcal{C}_4\)

b) \(^{14}\mathcal{C}_1\cdot^{10}\mathcal{C}_1\cdot^{13}\mathcal{C}_2\)

c) \(\dfrac{^{10}\mathcal{C}_2\cdot~ ^{22}\mathcal{C}_2}{^{24}\mathcal{C}_4}\)
 

Sonal7

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#9 a) \(^{24}\mathcal{C}_4\)

b) \(^{14}\mathcal{C}_1\cdot^{10}\mathcal{C}_1\cdot^{13}\mathcal{C}_2\)

c) \(\dfrac{^{10}\mathcal{C}_2\cdot~ ^{22}\mathcal{C}_2}{^{24}\mathcal{C}_4}\)
The answer for a) is 24C4 x4 x3. I am confused about the answer. I am glad that it matches my answer for (a). There are 2 treasurers so I think there is a repetition. That means that 4x3 x 25C4 is correct. as the two treasurers are the same but I am not sure exactly why. This is so hard :(
 
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Sonal7

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Oh i think its 4! % by 2! for the two treasurers.

The ans for c is also the same as yours but x2. So there is a permutation?
 
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pka

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The answer for a) is 24C4 x4 x3. I am confused about the answer. I am glad that it matches my answer for (a). There are 2 treasurers so I think there is a repetition. That means that 4x3 x 25C4 is correct. as the two treasurers are the same but I am not sure exactly why.
The part a) should have said: how many can the offices be filled, once the members of the committee are chosen?
If we have four people the four offices can be filled in \(\dfrac{4!}{2!}\) ways that in the number of ways to arrange the string \(CSTT\).
 

Sonal7

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The part a) should have said: how many can the offices be filled, once the members of the committee are chosen?
If we have four people the four offices can be filled in \(\dfrac{4!}{2!}\) ways that in the number of ways to arrange the string \(CSTT\).
Thank you that makes sense. I had also the thought the question was not clear enough.
 
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