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| Math Central | Quandaries & Queries |
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Question from Jeannine, a parent: When using numbers from 1 to 70 in sets of 4, how many combinations are there without having the same combination twice? |
Hi Jeannine,
Mathematicians call these combinations, in your case the number of ways of choosing 4 things from 70 things. The expression for the number of ways of choosing k things from n things is
nCk = n!/[k! (n - k)!]
where n! = n × (n - 1) × (n - 2) ×···× 2 × 1, Hence
70C4 = 70!/[4! (70 - 4)!] = 70!/[4! 66!] = (70 × 69 × 68 × 67)/(4 × 3 × 2 × 1) = 916,895
If the n and k are large the arithmetic can become onerous. In this case Google can help. Go to Google and type the phrase 70 choose 4 into the search window. Google will respond with 70 choose 4 = 916 895.
Penny
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