permutations (1 Viewer)

[ahdil33]

Well, i said Just pick any two boys, make them one group in the first row. So now there are 3 groups at the front. we can arrange the boys, 2! and the three groups, so 3!

In the last row, there are 4 groups, so 4! . Besides the two boys at the front and the one girl at the back, everyone else can be arranged any way so that's 5!

in total we have 2! x 3! x 4! x 5!

Problem is the two boys aren't necessarily sitting together in the group. They could be on opposite ends of the front row, so that's why the 2! x 3! isn't correct for the front row, I believe.

The problem with 4! is that means your organising 4 people in 4 different spots. But you're not. It's only one person who's sitting there, and there's 4 options she can take.

And the 5! is right for the last 5, but that doesn't make sense if you've put 4! above.

 

[Shadowdude]

I got 5760 and thought I was wrong til I realised that was the answer you meant.

I just thought of it like this. Let's deal with the conditions first. We have one girl who has 4 possible seats she can sit in, in the back row

4...

Then we have the boys. The first boy can sit into another 4 possible seats, in the front row

4 X 4 ...

Next is the other boy, who can sit in 3 other seats left in the front row...

4 X 4 X 3

Now all restrictions are dealt with, and you have 5 people who left who don't care where they are. That means 5! ways of organising these five.

So, 4 X 4 X 3 X 5! = 5760

That logic looks good.