A four digit PIN is made from the digits 0 to 9. How many more PINs are possible if digits may repeat than if they may not?
A4960 more PINs
B5040 more PINs
C10000 more PINs
D3960 more PINs
Answer & Solution
Correct answer: A. 4960 more PINs
1. Count the two cases separately and then subtract.
2. With repetition allowed there are ten choices for each of the four positions.
3. So the count is 10 multiplied by 10 multiplied by 10 multiplied by 10, which is 10 000.
4. Without repetition the first digit still has ten choices, but each later digit has one fewer.
5. So the count is 10 multiplied by 9 multiplied by 8 multiplied by 7.
6. Ten multiplied by 9 is 90, 90 multiplied by 8 is 720, and 720 multiplied by 7 is 5040.
7. The difference is 10 000 minus 5040, which is 4960.
8. The answer 5040 is the trap for quoting the no repetition count itself, 10 000 quotes the repetition count, and 3960 subtracts 6040 by slipping a digit while multiplying.
_Source: Siyavula Mathematics Grade 12 (Everything Maths, CC BY 4.0), Chapter 10: Probability_
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