<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] 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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