To select a committee of 3 people from 10, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ). Here, ( n = 10 ) and ( k = 3 ). This gives ( C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 ). Therefore, there are 120 ways to select a committee of 3 people from 10.
240
There are 8!/(4!*4!) = 70 ways.
There are 9*8*7 = 504 ways.
For this type of problem, order doesn't matter in which you select the number of people out of the certain group. We use combination to solve the problem.Some notes to know what is going on with this problem:• You want to form a committee of 2 teachers and 5 students to be formed from 7 teachers and 25 students • Then, you select 2 teachers out of 7 without repetition and without considering about the orders of the teachers.• Similarly, you select 5 students out out 25 without repetition and without considering about the orders of the students.Therefore, the solution is (25 choose 5)(7 choose 2) ways, which is equivalent to 1115730 ways to form such committee!
To choose a president, vice president, and secretary from a committee of 8 members, we can select the president in 8 ways. After selecting the president, 7 members remain for the vice president, and after that, 6 members are left for the secretary. Thus, the total number of ways to choose these three positions is (8 \times 7 \times 6 = 336).
240
There are 8!/(4!*4!) = 70 ways.
There are 9*8*7 = 504 ways.
There are 45360 ways.
For this type of problem, order doesn't matter in which you select the number of people out of the certain group. We use combination to solve the problem.Some notes to know what is going on with this problem:• You want to form a committee of 2 teachers and 5 students to be formed from 7 teachers and 25 students • Then, you select 2 teachers out of 7 without repetition and without considering about the orders of the teachers.• Similarly, you select 5 students out out 25 without repetition and without considering about the orders of the students.Therefore, the solution is (25 choose 5)(7 choose 2) ways, which is equivalent to 1115730 ways to form such committee!
Congressman Mike Thompson is a member of the House Committee on Ways and Means and the House Permanent Select Committee on Intelligence. Mike Thompson represents the state of California.
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.
add ore people
To choose a president, vice president, and secretary from a committee of 8 members, we can select the president in 8 ways. After selecting the president, 7 members remain for the vice president, and after that, 6 members are left for the secretary. Thus, the total number of ways to choose these three positions is (8 \times 7 \times 6 = 336).
house has 106
There are 14C8 = 14*13*12*11*10*9/(6*5*4*3*2*1) = 3003 ways.
18x17= 306 ways