<s>[INST] How many five digit numbers can be formed using digits $1,1,2,3,3,4,4$ 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.
 [/INST] 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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