Question

Use the digits 0-9, find out how many 4 digit numbers can be configured based on the stated conditions:

a. The number cannot start with zero and no digits can be repeated.
b. The number must begin and end with an odd digit. (Repeated digits are okay)
c. The number must at least 5000 and be divisible by 10. (Repeated digits are okay)
d. The number must be less than 3000 and must be even. No digits may be repeated in the last 3 digits. (That is, 2,234 would be okay but 2,334 would not be okay)

Answer 1

is there a e) ignore all the above rules ?

Answer 2

a. If the number cannot start with zero, there are 9P3 permutations that are not allowed. Therefore the answer to a is found from:
10P4 - 9P3
Can you work this out and post the result ?

Answer 3

Work out 10P4-9P3?

Answer 4

4,536?

Answer 5

\[10P4-9P3=\frac{10!}{6!}-\frac{9!}{6!}=(10\times 9\times 8\times 7)-(9\times 8\times 7)=9\times 8\times 7\times (10-1)\]

Answer 6

yes 4,536

Answer 7

Yes! 4,536 configurations is correct :)

Answer 8

How about B,C, and D?

Answer 9

how do I solve those? or is 4,536 the entire answer?

Answer 10

b. There are 5P2 permutations of the 5 odd numbers taken 2 at a time. And there are 5P2 permutations of the even numbers taken 2 at a time (taking 0 as an even number)
Therefore there are 5P2 * 5P2 configurations:
\[\frac{5!}{3!}\times \frac{5!}{3!}=5\times 4\times 5\times 4=400\]

Answer 11

Okay, I get that one.

Answer 12

can you help me through the next two? Rather then telling me the answer lol:)

Answer 13

c. The first digit must be 5, 6, 7, 8 or 9. There are 9P3 permutations of the remaining numbers taken 3 at a time. Combined with the 5 numbers allowed for the most significant digit this gives 5 * 9P3 configurations. However the least significant digit must be 0 to enable division by 10, disallowing 9 numbers. Therefore the allowable number of configurations is:\[\frac{5\times 9P3}{9}=\frac{5\times 9\times 8\times 7}{9}=?\]

Answer 14

280 configurations?

Answer 15

Same answer as I get!
d. Only 0, 1 or 2 is allowed for the most significant digit. There are 9P3 permutations of the remaining 9 numbers taken 3 at a time without repetitions.
Therefore the number of configurations is 3 * 9P3 =
\[3\times \frac{9!}{6!}=3\times 9\times 8\times 7=?\]

Answer 16

1,512?

Answer 17

That's it! 1,512 configurations.

Answer 18

thank you!

Answer 19

You're welcome :)