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a.50 c.70 c 90 d 110

2006-08-19 07:43:20 · 3 answers · asked by guitarkhongday 1 in Science & Mathematics Mathematics

3 answers

6!/2!2!2! = 90

You are partitioning the 6 positions in the number into 3 groups of 2.
A group to hold the 2's a group to hold the 5's and a group to hold the 7's. This number is what is known as a multinomial coefficient.

2006-08-19 07:48:22 · answer #1 · answered by rt11guru 6 · 0 0

6 number in 6! ways but because there are repetitions, discount this by reptitions.

So, total number of combinations possible - 6!/2!2!2! = 90

2006-08-20 10:54:01 · answer #2 · answered by DG 3 · 0 0

I am just guessing.... since there are 3 different numbers I expect that the solutions must be divisible by 3 thus I choose 90

2006-08-19 14:51:44 · answer #3 · answered by gjmb1960 7 · 0 0

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