Math Questions [Combination ]

1) The 1st ball, you can put into any of the 4 boxes, so there are 4 ways. The 2nd ball, you can put into any of the 4 boxes again since there is no restrictions, similar for the 3rd, 4th and 5th ball, so the number of possible arrangement is 4 x 4 x 4 x 4 x 4

2) The 1st passenger can enter any of the 8 coaches, so there are 8 ways. The 2nd passenger cannot enter the same coach, so he has only 7 ways. So the answer is 8 x 7

3) Same reasoning as 2) so the answer is 4 x 3

4a) The 1st woman can choose any men, so she has 4 choices. The 2nd woman has only 3 and so forth, so the answer is 4 x 3 x 2 x 1
4b) This is a bit tricky. Think of it like this. There is only 1 way for everyone to have his wife as partner. If this is not the case, then two will not have their wives as partner because there cannot be only one does not have his wife. So the answer is (4 x 3 x 2 x 1) - 1

5) Since no 3 points are collinear, so any 2 points will form a line, so the answer is just C(10, 2) = (10 x 9)/2

6) May is always in the group, so the answer is pick 3 teacher from the 7 remaining teachers. So the answer is C(7,3) = (7 x 6 x 5)/3!

The answer of 4(b) should be 1.
Only 1 way : 2 couples as partners, and the other 2 couples change their partners.