I think in this case first we select and then arrange
in your example first we select 5 people from 7 in 7C5 ways
then arrange them in (5-1)! ways

hence total ways are (7C5)4! ways

hence total ways are (7C5)4! ways

Are you saying (7C5)*4!, if so what does 7C5 equal? I'm not familar with this terminology. Thanks for the relpy!

is it.....???

7C5 = 7!/(5!*2!) = 21

therefore...

21*4! = 504

I would be helpful if someone could confirm this. thanks.

scoowhoop wrote:

hence total ways are (7C5)4! ways

Are you saying (7C5)*4!, if so what does 7C5 equal? I'm not familar with this terminology. Thanks for the relpy!

Hi,

7C5 represents the combinations formula; you say it out loud as "seven choose 5".

We use the combinations formula when we're counting unordered subgroups. The general formula is:

nCk = n!/k!(n-k)!,

in which:

n = total number of objects; and
k = # of objects that we're selecting.

In the question you posted, we have 7 total people and we're selecting 5, so n=7 and k=5.

7C5 = 7!/5!(7-5)! = 7!/5!2! = 7*6/2*1 = 42/2 = 21

In other words, if you have 7 distinct objects, you can select 21 different groups of 5 of them.

Thanks Stuart!

So just to clarify the answer would be 504, correct?

scoowhoop wrote:Thanks Stuart!

So just to clarify the answer would be 504, correct?

Correct!