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This is known as the number of combinations of 25 things taken 6 at a time. It is equal to 25!/(6!19!) where n! is n factorial which means the product of the integers from 1 to n. This works out to 20 * 21 * 22 * 23 * 24 * 25 / 1 * 2 * 3 * 4 * 5 * 6. After canceling we are left with 10 * 7 * 22 * 23 * 5. I get 177,100

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Q: How many ways can a 6 person committee be selected from a group of 25?

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-1

There are 8!/(4!*4!) = 70 ways.

Nine people can be selected from the group 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 is79,833,600/362,880 = 220 .

30240

Typically a group of people selected from within a corporation that will work together in a committee to make an event happen. this will include budgeting, promoting and scheduling along with many other functions.

3 people can be selected from a pool of 7 people in (7 x 6 x 5) = 210 ways.But each group of 3 can be selected and seated in (3 x 2 x 1) = 6 ways.So the number of different 3-person subcommittees formed from 7 people is (210/6) = 35 .

There is no specific number of vultures in a committee. A committee of vultures is just a group of vultures.

5 for 2, 3 for 3, 2 for 4.

The number is 7C3 = 7*6*5/(3*2*1) = 35

-5

18x17= 306 ways

53,130 ways.

560.

4 people can be selected from a pool of 15 people in (15 x 14 x 13 x 12) = 32,760 ways.But each group of 4 can be selected and seated in (4 x 3 x 2 x 1) = 24 ways.So the number of different 4-person subcommittees formed from 15 people is (32,760 / 4) = 8,190 .

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 is79,833,600/362,880 = 220 .

One.

7C4 = 35

15

12C9 = 220

There are 14C8 = 14*13*12*11*10*9/(6*5*4*3*2*1) = 3003 ways.

35

The short notation of this would be 12C2 There are 12 possible selections for the first person, and 11 possible selections for the second person. 12x11 is 132 But half of these options are the same as another option but with the people being picked the other way around, so for the real value we need to do 132/2 or 66. Thus, there are 66 ways a committee of 2 can be selected from a club with 12 members.

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