How many different combinations of 3 can you make out of the 9 side dishes? The side dishes are salad, cole slaw, beans, fruit, corn, mac & cheese, potato, biscuit, and watermelon.
Show all work and explain how you did it. Is there an easy way to do it?
2006-11-20 07:26:58 · 5 answers · asked by Katey 1 in Science & Mathematics ➔ Mathematics
You have 9 choices for the first dish, 8 for the second dish and 7 choices for the third.
This would be 9 x 8 x 7, but you end up repeating combinations because order doesn't matter. For example salad, beans and fruit is the same as fruit, salad and beans. So we need to divide by the number of ways to arrange 3 items. This is 3! (3 x 2 x 1).
So the final answer is: (9 x 8 x 7) / (3 x 2 x 1) = 84 ways
This is used so often that we it has a specially name of "n choose k", being the number of ways to choose k items from n total items. It is either written nCk or C(n, k)
The formula is:
C(n, k) = n! / (n-k!)k!
In your example:
C(9, 3) = 9! / (9-3)!3!
C(9, 3) = 9! / 6!3!
C(9, 3) = 84
(Note: if you have a computer handy, you can type "9 choose 3" into Google it will even give you the answer.)
2006-11-20 07:34:44 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
There are 9 ways of choosing the first, 8 ways of choosing the second and 7 ways of choosing the third.
So there are 9 x 8 x 7 ways of choosing 3 dishes out of nine,
It does not matter in what order you choose the dishes (the word combination is a clue here).
Within those three dishes they can be arranged 3 x 2 x 1 ways
To get the number of combinations you divide 9 x 8 x 7 by 3 x 2 x 1
=3 x 4 x 7
= 84 different different combinations
2006-11-20 07:46:43 · answer #2 · answered by rosie recipe 7 · 0⤊ 0⤋
3 x (9+8+7+6+5+4+3+2+1) =
3 dishes times the number of available dishes.
2006-11-20 07:55:21 · answer #3 · answered by scottboss64 3 · 0⤊ 0⤋
9*8*7/(3*2*1)=84
2006-11-20 08:03:08 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
9C3
=9!/3!*6!
=84
2006-11-20 07:28:56 · answer #5 · answered by Dupinder jeet kaur k 2 · 1⤊ 0⤋