# 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.

• Yes your working is correct (after the edit you made). Unfortunately, the correct answer is not there in the options. Mar 26 at 14:51
• Thank you for pointing that out! Mar 26 at 14:51

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.

• I just fixed my mistake in case one so my answer is now $5!\times3$. and yes now we have the same answer! Mar 26 at 14:49
• Ok updated to reflect that Mar 26 at 14:50
• Oh ok, thank you very much! Mar 26 at 14:51

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, $$\frac{5!3}{2!2!} + \frac{5!3}{2!2!} +\frac{5!3}{2!} = 5!3$$