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bezimeni [28]
3 years ago
9

In Exercise #13 above, how many different combinations would be possible if the three numbers do not have to be different (for e

xample, 20-20-20 could be a combination)?
Mathematics
1 answer:
Ainat [17]3 years ago
8 0

Answer:

 116 280

Step-by-step explanation:

number of combination = 20 x 19 x 18 x 17 = 116 280

HOPE THIS HELPED ;3

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