Solution:
The number of methods can be calculated using combinatorial formulas:
10! /(2!3!5!)
10!/(2!3!5!) total number of permutations;
5⋅2 ways that on one of the silver lies 3 red and 2greens can be placed in different ways;
5 ⋅2 the ways that one of the silver ones has two red ones and 2 greens can be placed in different ways;
25 ⋅ 2 ways that one lies 2 red and the other 1 red and 2 greens can be placed in different ways;
10! /(2!⋅3!⋅ 5!) -10-10-50= 2090 ways;
Answer: 2090 ways.
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