How many ways are there to select a subcommittee of 7 members from among a committee of 33?

Answer 1

Combination of 7 out of possible 33 which is [33!/(7! x 26!)] = 27.28.29.30.31.32.33/1.2.3.4.5.6.7 which is 27.28.29.31.32,33/1.2.3.4.7 = 27.29.31.32.33/6 = 9.29.31.32.33/2 = 9.29.31.16.33 = 4272048 ways

Answer 2

The first person selected can be any one of 33. For each of those . . .
The second person selected can be any one of the remaining 32. For each of those . . .
The third person selected can be any one of the remaining 31. For each of those . . .
The fourth person selected can be any one of the remaining 30. For each of those . . .
The fifth person selected can be any one of the remaining 29. For each of those . . .
The sixth person selected can be any one of the remaining 28. For each of those . . .
The seventh person selected can be any one of the remaining 27.

So the number of ways to select 7 out of the 33 is (33 x 32 x 31 x 30 x 29 x 28 x 27) = 2.153 x 10^10.

But each group of 7 can be seated in (7 x 6 x 5 x 4 x 3 x 2 x 1) = 5,040 different ways.
So there are 5,040 different ways to select each 7-member subcommittee.

So the number of different 7-member subcommittees that can be selected from 33 people is

2.153 x 10^10 / 5,040 = (33 !) / (26 !) x (7 !) = only a mere 4,272,048 .

Answer 3

33 over 7 (written without a fraction line, and with parentheses around the entire expression); this is short for: (33 x 32 x 31 x 30 x 29 x 28 x 27) / (1 x 2 x 3 x 4 x 5 x 6 x 7).

Answer 4

The answer is 33C7 = 33 !/[7!*(33-7)!] = 4,272,048 ways.