it is a combination: 9!/4!=9 x 8 x 7 x 6
There are 9*8*7 = 504 ways.
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
40 x 39 x 38 x 37 = 2193360
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.
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.
There are 11880 ways.