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

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)