Number of 5 digit numbers $< 40,000$

The numbers to be used are : 2, 3, 4, 4, 5

The way I approached this is:

Total number of combinations possible is :

$$\frac{5!}{2!}$$

Total number of combinations starting with 4 :

$$4!$$

Total number of combinations starting with 5 :

$$\frac{4!}{2!}$$

$\therefore$ the total number of numbers $<$ 40,000 :

$$\frac{5!}{2!}-\Big(4!+\frac{4!}{2!}\Big)$$

I came across with this question and I don't have access to the solution.

I'm not confident if this is correct.

• You must use the digits $2,3,4,4,5$ ? – user252450 Dec 17 '15 at 7:56
• The solution is correct .Try to look at it again and understand why it's correct . – user252450 Dec 17 '15 at 8:01
• @ComplexPhi Yes we must use those digits only – Siddharth Thevaril Dec 17 '15 at 8:04

You have to use digits $2,3,4,4,5$, and it can't exceed $40000$. That means, you can't start your number with $4$ or $5$. That leaves you with the option to start with $2$ or $3$. If you started your number, you still have $4$ digits, but $2$ of them are the same, so the total possibilities are $\frac{4!}{2!}$, and you multiply this with $2$, since you can start with either $2$ or $3$, therefore the answer is $24$. Your solution is correct, but I just wanted to show you a "clearer" one.
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