Probability Q-combinatorics type

[Sonal7]

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]

#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]

#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

 

[Sonal7]

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?

 

[pka]

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]

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.