Here is an exercise I could not solve:

A = Number of 4-digit numbers, which do not have two equal consecutive digits.

B = Number of 4-digit numbers, which do not have two equal digits.

How much is A - B?

I translated the exercise, so it may sound a litter weird to you. Here is what I have done:

If I understood the condition for B correctly, only the 4-digit numbers that do not have repeated digits satisfy the condition, so $B = 9\times9\times8\times7$.

A includes B but also numbers that do have repeated numbers, just not one next to the other, something like $abca$ $abac$ $baca$, so $A = B + 9\times9\times8\times3$.

Which would mean $A-B = 9\times9\times8\times3$, which is incorrect because does not correspond to any of the four options for the answer.

I apologize in advance for any dumb errors. I have always done terribly at combinatorics.

  • $\begingroup$ The first digit must be 1-9 and cannot be zero. Consider A: there are 9 possible digits, 1-9, whatever digit you choose for the 2nd, it cannot be the one you chose for the 1st, so there are 9 digits (because there are 8 plus it can now be zero), there are 9 possible digits for the 3rd (because it can be anything but the previous--including 0), and 9, again, for the 4th. For B: again there are 9 possible digits for the 1st, then 9 again for the second (again 8 of 1-9 plus zero), then 8 for the 3rd (you've eliminated 2 of 10 possible digits), and finally 7 for the 4th. $\endgroup$ – Jared Feb 19 '16 at 5:14
  • $\begingroup$ Since you are going to subtract A from B anyway, don't try to calculate A around B. That's just too hard. A. There are 9 choice for the first digit,9 for the second etc. So A is 9^4. B is 9x9x8x7. So A - B = 9x9x (81-56) $\endgroup$ – fleablood Feb 19 '16 at 6:40

A: you can have 9 digits for the first digit. (Anything but zero) You have 9 choices for the second digit (anything but the first digit). You have 9 for the third (anything but the second) and nine four the 4th (anything but the 3rd.)

So A =$9^4$.

B)no repeats. 9 for the first (anything but 0). 9 for the second (anything but the first). 8 for the 3rd (anything but the first and second). And 7 for the 3rd.

So B = $9*9*8*7$

A-B = $9*9*(81-56)=81*25=2025$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.