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Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. What appeared instead was the five-digit number 52115. How many different seven-digit numbers could Sid have meant to type?A. 10B. 16C. 21D. 24E. 27

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abidemiokin

Answer: D. 21ways.

Step-by-step explanation:

This is based on selection, fixing two '3' numbers into 5numbers to make it 7 numbers can be done in 7C2 ways.

Generally, nCr = n!/(n-r)!r!

7C2 = 7!/(7-2)!2!

=7!/5!2!

=7×6×5!/5!×2!

=7×6/2

=21ways (D)

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