2
$\begingroup$

How many distinct four-digit integers can one make from the digits $1$, $3$, $3$, $7$, $7$ and $8$?

I can't really think how to get started with this, the only way I think might work would be to go through all the cases. For instance, two $3$'s and two $7$'s as one case, one $1$, two $3$'s and one $8$ as another. This seems a bit tedious though (especially for a larger alphabet) and so I'm here to ask if there's a better way.

Thanks.

$\endgroup$
  • $\begingroup$ I don't see any other simpler method. $\endgroup$ – robjohn Jun 2 '12 at 7:58
  • $\begingroup$ Thanks for the answer, you simplified it a bit from what I was thinking by focusing on the 7s and the 3s. Cheers. $\endgroup$ – urbanyoung Jun 2 '12 at 8:19
3
$\begingroup$

Distinct numbers with two $3$s and two $7$s: $\binom{4}{2}=6$.

Distinct numbers with two $3$s and one or fewer $7$s: $\binom{4}{2}3\cdot2=36$.

Distinct numbers with two $7$s and one or fewer $3$s: $\binom{4}{2}3\cdot2=36$.

Distinct numbers with one or fewer $7$s and one or fewer $3$s: $4\cdot3\cdot2\cdot1=24$.

Total: $6+36+36+24=102$


With larger alphabets,

Suppose there are $a$ numbers with 4 or more in the list, $b$ numbers with exactly 3 in the list, $c$ numbers with exactly 2 in the list, and $d$ numbers with exactly 1 in the list.

Distinct numbers with all 4 digits the same: $a$

Distinct numbers with 3 digits the same: $\binom{4}{3}(a+b)(a+b+c+d-1)$

Distinct numbers with 2 pairs of digits: $\binom{4}{2}\binom{a+b+c}{2}$

Distinct numbers with exactly 1 pair of digits: $\binom{4}{2}(a+b+c)\binom{a+b+c+d-1}{2}2!$

Distinct numbers with no pair of digits: $\binom{a+b+c+d}{4}4!$

Total: $a+4(a+b)(a+b+c+d-1)+6\binom{a+b+c}{2}+12(a+b+c)\binom{a+b+c+d-1}{2}+24\binom{a+b+c+d}{4}$

Apply to the previous case: $a=b=0$, $c=2$, and $d=2$:

$0+0+6\binom{2}{2}+12(2)\binom{3}{2}+24\binom{4}{4}=102$

$\endgroup$

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.