A photographer is to take group photographs of a class of students for

I went down the wrong path thinking I'd calculate the total possibilities and subtract out the possibilities where 2 Boys come together or 2 Girls come together. I stumbled upon the right answer anyway because of ball parking, In hindsight I'd suggest using the methods shown above but I'll share what I did just in case it helps.

so first off, 4 G and 4 B. So total 8 people.

Selecting 5 from 8 and where arrangement matters, is \(8P5\) or\( 8C5! * 5!\)
= \(\frac{8!}{3!} = 8*7*6*5*4= 40 * 7 * 24 = 280 * 24\)

Now we know we need to remove the occurrences where BB _ _ _ or GG _ _ _ or both BB GG _
First case, \(2C1 * 4C2 * 4! \)(Selecting either B or G , Selecting 2 from 4 , arranging the grouped 2 , and the rest)
\(=2* 6 * 120 = 1440\)
The other case will definitely be significantly lower than this.

Now looking at the options I Know Option A, C and E are out. Judging by the magnitude, I'd also fairly assume Option B is out and came to the solution of Option D.


Of course, none of these issues would occur if I had not gone down the wrong rabbit hole but hey, here's a not so bad to get out of it in time.

The question asks how many different photographs can the photographer take, so why is the answer only accounting only for the first photograph taken? 576 is the number of ways of taking the first photo.

But we are still left with 3 students from the class, which will be either GBG or BGB. Why aren't we accounting for that?

chetan2u if you can advise.

No, the question is how many photographs that have 5 students. So these combinations of 3 left over students is out.

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