Nine people can be selected from the group of 12 in
(12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4) = 79,833,600ways.
But each group of the same 9 people can be selected in
(9 x 8 x 7 x 6 x 5 x 4 x 3 x 2) = 362,880 different orders.
So the number of different 9-person committees that can be selected is
79,833,600/362,880 = 220 .
There are 8 ways to choose the first book There are 7 ways to choose the second book - 8 x 7 = 56 ways to select two books There are 6 ways to choose the third book - 8 x 7 x 6 = 336 way to select three books There are 5 ways to choose the fourth book - 8 x 7 x 6 x 5 = 1,680 ways to select four books.
For the first spot, you can choose any one of 5 students. For the second spot, you can choose any one of the remaining 4 students. For the third spot, you can choose any one of the remaining 3 students. etc. So the answer is: 5x4x3x2x1 = 120
Combination of three people from a group of 10 = (n!)/[r! x (n-r)!], where n = 10 and r = 3. The answer is therefore (10.9.8.7.6.5.4.3.2.1)/[(1.2.3) x (1.2.3.4.5.6.7)] = (10.9.8)/(1.2.3) = 10.3.4 = 120 ways
I Dont Know [210]
The answer is 10 over 3 (you write 10 over 3, without a fraction line in between, and with parentheses around the entire expression). This is calculated as (10 x 9 x 8) / (1 x 2 x 3).
7C4 = 35
12C9 = 220
Any 4 from 7 = 35 ways
Well, honey, if you want a committee with at least 3 educators out of 5 members, you can choose 3 educators out of 8 in 8 choose 3 ways, and then fill the remaining 2 spots with any of the remaining educators or accountants. So, that gives you 8 choose 3 * 5 choose 2 ways, which is 56 * 10 = 560 ways to create your sassy little committee.
7*6*5*4/(4*3*2*1) = 35 ways
7
-1
There are 45360 ways.
40 x 39 x 38 x 37 = 2193360
house has 106
There are 14C8 = 14*13*12*11*10*9/(6*5*4*3*2*1) = 3003 ways.
18x17= 306 ways