1 Answer. There are 20 ways to choose 3 students from a group of 6 students.

3! (10−3)! = 120.

= 25! 3! (22)! So, there are 2300 different committees that can be formed.

So substituting 12 for n and 5 for r in the above equation yields a value of 792 possible ways of choosing 5 people at a time from a set of 12 people.

20–1)! = 24! / 5!* 19! The answer is therefore that you can create 42504 unique committees of 5 people from 20 different people.

SOLUTION: How many different committees of 7 people can be formed from a group of 10 people. But, choosing A,B,C,D,E,F,G is the same as A,C,D,E,F,G,B. There are 7*6*5*4*3*2*1 ways to choose any group of 7, = 5040 ways. 604800/5040 = 120 different committees.

There are 6 different choices for the fourth person. This gives 9×8×7×6 different committees, however this will include the same combinations of people. There are 4×3×2×1 ways in which 4 people can be chosen. 9×8×7×6×54×3×2×1=126 different committees.

Therefore, in total, there are 315 + 210 + 35 = 560 possible 3-person committees.

5C3 ways

There are 4C1*6=24 ways there to be a couple among 3 members: 4C1 ways to select a couple out of 4, which will be in the committee and 6 ways to select the third remaining member (since there will be 6 members left after we select a couple out of 8 people). 56-24=32.

1320 different ways

Question 691107: How many ways can a president, vice-president, and secretary be chosen from a committee of 7 people? 7*6*5=210 WAYS TO SELECT THESE 3 POSITIONS.

126 different

In all, there are 12*11*10*9 = 11880 ways.

in how many ways can a president, vice president, and a secretary be chosen?… is it 12X11X10. Permutation of n things taken r at a time: nPr=n!/(n-r)! 12P3=12*11*10*9!/9!= 12*11*10=1320 ways.

A group of 4 people are standing in a straight line. In how many different ways can these people be standing on the line? The answer is 24.

Therefore, the 3 groups can be chosen 84 x 20 x 1 = 1680 ways. However, since the order of the 3 groups doesn’t matter, we have to divide 1680 by 3!. Hence, the number of ways 9 people can be divided into 3 groups is 1680/3! = 1680/6 = 280.

If you meant to say “permutations”, then you are probably asking the question “how many different ways can I arrange the order of four numbers?” The answer to this question (which you got right) is 24. Here’s how to observe this: 1.

40,320 different combinations

127

1,023

Team of any 5 numbers can be chosen from 50 numbers in (50C5) combinations. Now, we are to choose 10 numbers from the original pool of 50 numbers such that all previous ‘five-number combinations’ are covered.

In your case, with 12 numbers, the number is 12x11x10x… x2x1=479001600. This number is called “twelve factorial” and written 12!, so, for example 4!= 4x3x2x1=24.