Four married couples have bought 8 seats in the same row for a concert. in how many different ways can they be seated (a) with no restrictions? (b) if each couple is to sit together?

(a)

Start at 8 and count your way down until you get to 1. Multiply all the values out. This is using a factorial

8! = 8*7*6*5*4*3*2*1 = 40320

This works because we have 8 choices for slot A, then 7 for slot B, and so on until we get down to just one choice for slot H

(b)

If we must make restrictions that the couples sit together, then we can think of each couple as a "person". Consider person A and person B are married together, the couple can be labeled "person C". Treating this couple as one person allows us to move the blocks of four couples to arrange them into 24 different ways. Where is the 24 coming from, well,

4! = 4*3*2*1 = 24

which is similar to what happened in the previous part above. Now we start with 4 instead of 8.

So if we had "person"s A, B, C, D representing the four couples, then there are 24 ways to arrange the "person"s, or 24 ways to arrange the couples.

Then looking at each block, there are 2 ways to arrange any one couple. So there are 2*2*2*2 = 16 ways to arrange all the couples if they stay in their same block (swap the two people for any single block). Therefore we multiply 24 by 16 to get 24*16 = 384

Side note: for either (a) or (b) you can use the permutation formula

nPr = (n!)/(n-r)!

For part (a), use n = 8 and r = 8 which effectively simplifies to n! = 8! = 40320. Part (b) will have n = 4 and r = 4 to get 24, which you'll multiply by 16 to get 384.

found it xd it ok yt

1/800 is the right answer