How any people are there in a class in which there are 20 ways to select a committee of three people?

Question

I really get stuck on this problem
All I know that I can get the answer by solving this equation
nC3=20

How many people are there in a class

Answer 1

If you have n people in the class, then you have n choices for the first seat in the 3-seat committee; this leaves (n-1) choices for the second seat, and (n-2 for the third seat. So the total number of possible combinations are
(n)(n-1)(n-2)
However, if you pick person A for seat 1, person B for seat 2, and person C for seat 3, then you get the same committee as if you picked A,C,B, or B,A, C, or B, C, A, or C, A, B, or C, B, A. You've overcounted by 3!, or (3*2*1), which is 6. So you have to reduce your possible combinations by 3!, leaving
(n)(n-1)(n-2)/3!
Now try some numbers for n. If n=3, then you get (3*2*1)/6, or 1 way to fill the three seats. If n=4, then you get (4*3*2)/6, or 4 ways. If n=5, you get (5*4*3)/6, or 10 ways.

Now try n=6 and I think you'll be pleasantly surprised.

Answer 2

There ar 6 people in the class.

Number of ways to pick 3 out of 6 is 6*5*4/1*2*3 =20

In general, for a class of N people the number of ways you can choose a commmitee of K people is
N*(N-1)*(N-2)*...(N-K+1) / 1*2*3...*K

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Answer 3

Neither of the first two people answered your question, and both their answers had flaws.

You are right.

Using the definition of nC3:

n!/(n-3)! = 3!*20 = 120 ( which is nP3, permutations)

Now, you can go all crazy with n, n-1, n-2 ( like Renny B who wants you to fool around with a third order polynomial!!)
or you can just try a few small values:

Table:
n, nP3
-----------
6, 6*5*4 = 120, <- *** ding, ding, ding *** first try!!!

So, a check:
6C3=20 = 6*5*4/(3*2) = 20

Answer 4

nC3 = n!/(3! * (n - 3)!) = 20
n! = 20 * 6 * (n - 3)!

divide both sides by (n - 3)!

n * (n-1) * (n-2) = 120

I'll leave the rest to you!

(Everyone else - read the question! There aren't 20 people to pick from, there's 20 possible combinations!)

Answer 5

You are right about the nC3=20
Remember that the formula for nCr= n!/(n-r!)*(r!)
So in this case you would have this:

n!/(n-3)!(3!)=20 now expand n! so you can divide to n(n-1)(n-2)(n-3!)
n(n-1)(n-2)/6=20
n(n-1)(n-2)=120
n^3-3n^2+2n=120
n^3-3n^2+2n-120=0

from there you solve and find that n=6

Answer 6

okay so if you pick 3 people, you obviously cant pick the same person twice. so the first time you pick someone from the group there are 20 different possibilities. the second time you pick someone for the group there are 19 possiblities (you already picked one), and the third time you have 18 possibilities (you already picked two).

therefore, you can find the number by multiplying them, because these events are happening AT THE SAME TIME

thus you get 20*19*18, or 20P3
= 6840 possibilities

Answer 7

but you can have two people the same and change out the third, right? Or have combos therein?