Can anyone help with number 54?
1 answer:
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There are ways of picking 2 of the 10 available positions for a 0. 8 positions remain.
There are ways of picking 3 of the 8 available positions for a 1. 5 positions remain, but we're filling all of them with 2s, and there's
way of doing that.
So we have
The last expression has a more compact form in terms of the so-called multinomial coefficient,
14, 15, 20 | 2
7, 15, 10 | 2
7, 15, 5 | 3
7, 5, 5 | 5
7, 1, 1 | 7
1, 1, 1
Answer: × 3 × 5 × 7 = 420
Hope it helped,
Happy homework/ study/ exam!