it is a combination: 9!/4!=9 x 8 x 7 x 6
There are 9*8*7 = 504 ways.
You can choose 4 people in (17 x 16 x 15 x 14) = 57,120 ways.But there are (4 x 3 x 2) = 24 ways to pick the same 4 people, so there are only57,120/24 = 2,380 different groups of 4 people that you can end up with.
The number of ways is 18C5 = 18!/(5!*13!) = 8,568 ways.
First in line can be any of the 12, second can be any of the remaining 11, third any of 10 and fourth any of 9 so 12 x 11 x 10 x 9 ie 11880 different ways
It is: 15C7 = 6435 combinations
it is a combination: 9!/4!=9 x 8 x 7 x 6
20C2 = 190
There are 9*8*7 = 504 ways.
The answer is 15 * 14 * 13 = 15 P 3 = 780 IF you assume that no one person can hold two offices at once and that all in the group are qualified for any office.
You can choose 4 people in (17 x 16 x 15 x 14) = 57,120 ways.But there are (4 x 3 x 2) = 24 ways to pick the same 4 people, so there are only57,120/24 = 2,380 different groups of 4 people that you can end up with.
The number of ways is 18C5 = 18!/(5!*13!) = 8,568 ways.
3 students can be chosen from a class of 30 in (30 x 29 x 28) = 24,360 ways.But each group of the same 3 students will be chosen in six different ways.The number of different groups of 3 is 24,360/6 = 4,060 .
First in line can be any of the 12, second can be any of the remaining 11, third any of 10 and fourth any of 9 so 12 x 11 x 10 x 9 ie 11880 different ways
40 x 39 x 38 x 37 = 2193360
10
I'm going with 25,200 3 men out of 10 may be chosen in 10C3 ways = 10 ! / 3! 7 ! = 120 ways. 4 women may be chosen out of 10 in 10C4 = 10 ! / 4! 6! ways = 210 ways. Therefore, a committee with 3 men and 4 women can be formed in 120 x 210 = 25,200 ways.