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With digits $1,1,2,3,3,4,4$, how many five digit numbers we can form?

$1)\frac34\times5!\qquad\qquad2)\frac94\times5!\qquad\qquad3)4\times5!\qquad\qquad4)\frac52\times6!$

Ok so the digits $1,3,4$ appears twice and $2$ appears once. I tried to count different cases:

first: having all $1,2,3,4$ digits and choose another digit from $1,3,4$: $${3\choose1}\times\frac{5!}{2!}$$ second: having two pair of same digit and choose another digit: $${3\choose2}\times2\times\frac{5!}{2!2!}$$ Summing them we have $3\times\dfrac{5!}{2}+3\times\dfrac{5!}{2}=5!\times3$ but I don't have this in the options.

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  • $\begingroup$ Yes your working is correct (after the edit you made). Unfortunately, the correct answer is not there in the options. $\endgroup$
    – Math Lover
    Commented Mar 26, 2021 at 14:51
  • $\begingroup$ Thank you for pointing that out! $\endgroup$
    – Amirali
    Commented Mar 26, 2021 at 14:51

2 Answers 2

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We have two multiplicity patterns possible for numbers given the digits available:
$(2,1,1,1) \approx wwxyz$ (A) and $(2,2,1)\approx wwxxy$ (B).

Pattern A can be filled in $\binom 31 \binom 33 = 3$ ways and permuted in $\binom{5}{2,1,1,1}$ $ = 5!/2! = 60$ ways

Pattern B can be filled in $\binom 32 \binom 21 = 6$ ways and permuted in $\binom{5}{2,2,1} = 5!/(2!2!) = 30$ ways

So $180+180 = 360$ options altogether which matches your revised result but again doesn't match any given options.

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  • $\begingroup$ I just fixed my mistake in case one so my answer is now $5!\times3$. and yes now we have the same answer! $\endgroup$
    – Amirali
    Commented Mar 26, 2021 at 14:49
  • $\begingroup$ Ok updated to reflect that $\endgroup$
    – Joffan
    Commented Mar 26, 2021 at 14:50
  • $\begingroup$ Oh ok, thank you very much! $\endgroup$
    – Amirali
    Commented Mar 26, 2021 at 14:51
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Either you don't select $2$ or you select $2$. In the first case you have 3 choices: $(2, 2, 1); (2, 1, 2); (1, 2, 2)$. For each such choice you have $\frac{5!}{2!2!}$ allocations. If you select $2$, you have two paths:$(2, 2, 0), (2, 0, 2), (0,2,2)$, in which case you have $\frac{5!}{2!2!}$ allocations and $(1, 1, 2), (1,2,1),(1,1,2)$, in which case you have $\frac{5!}{2!}$ allocations. Putting it together, $$ \frac{5!3}{2!2!} + \frac{5!3}{2!2!} +\frac{5!3}{2!} = 5!3 $$

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