Can you help me figure out the formula to solve this problem?

Question

I have 4 items. I want to find out how many combinations of 2 items I can make without any duplication. For example, A and B is the same as B and A. I know I can just write out the combinations but if I have, say 100 items, this would be time-consuming.

Ah, yes permutations! I knew it had a name but I couldn't remember. Thanks all! I'll choose a best answer as soon as Yahoo will allow me.

Answer 1

This is sampling without replacement and without regard to order.

It's given by the binomial coefficient:

for n items, choosing k at a time

nCk means n choose k

or

n!
----------- (this is a division)
(n-k)!k!
Note that you can get these from Pascal's triangle, nth row, kth column.

So, 4C2 is 6:

AB
BC
CD
DA
AC
BD

Answer 2

it is 4C2 or combination of 4 takentwo at a time
=4*3/1*2=6

if you have 100 items and you want tio find the combinations taking 90 at a time it will be 100C90
=100!/10!(100-10)!

Answer 3

The formula for the number of combinations of n different things taking r things at a time (when r is less than or equal to n) is
n!/(r!){(n-r)!},where n! means permutation of n,r! means permutation of r and (n-r)! means permutation of (n-r)

Hence your answer for 100 different things taken 2 at a time is
100!/2!98! But as you know 100! is 100X99X98X97......X3X2X1 and 98! is 98X97........X3X2X1,we have to calculat only for
100X99/2 and our answer would be 50X99=4950

Answer 4

4C2 = (4!)/(2!(4 - 2)!)
4C2 = (4!)/(2 * 2!)
4C2 = (4!)/(4)
4C2 = 3!
4C2 = 6

-------------------------

100C2 = (100!)/(2!(100 - 2)!)
100C2 = (100!)/(2 * 98!)
100C2 = 4950

For more info, go to http://mathforum.org/dr.math/faq/faq.comb.perm.html