If she will choose 8 from 12 photos, the total number of ways she can choose is given by a combination of 12 choose 8, since the order of the photos doesn't matter.

The formula for a combination of n choose p is:

$C(n,p)=\frac{n!}{p!(n-p)!}$

For n = 12 and p = 8, we have:

$C(12,8)=\frac{12!}{8!(12-8)!_{}}=\frac{12\cdot11\cdot10\cdot9\cdot8!}{8!\cdot4!}=\frac{12\cdot11\cdot10\cdot9}{4\cdot3\cdot2}=495$

So there are 495 ways.

If she wants to arrange the 12 photos, the total number of ways is given by the factorial of 12:

$12!=12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2=479001600$

There are 479,001,600 ways.

Since 10 photos already have specific places, we need to calculate the number of ways to arrange the other two photos in the two remaining places.

In this case, there are only 2 ways of organizing the remaining two photos:

Photo 1 first, photo 2 last, or photo 1 last and photo 2 first.

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1 year ago

Hi anyone bored here........​

Hi anyone bored here........