Permutation of objects selected from a group

Question

I'm having trouble with this question:

How many numbers greater than 300 can be formed from the figures 4, 3, 2 and 1 if each figure can be used no more than once in each number and all the figures need not be used each time?

Here is my approach. Considering that the first digit can either be 3 or 4, this leaves us with 2 options for the first digit. The count of all possible 3-digit numbers (i.e. 321) would be:

$$2 × 3! × 2!$$

And the count for all possible 4-digit numbers would be:

$$2 × 3! × 2! × 1!$$

So the total would be $2 × 24 = 48$. But the answer listed is 36. Where am I going wrong?

Answer 1

First of all, you should remove factorials the in your expressions (do you see why?). Second, the first digit in the 4-digit number does not have to be 3 or 4.

Answer 2

your difficulty is with the 3-digit numbers. here the first digit must, as you say, be 3 or 4. for the remaining two digits there are $\binom{3}{2}=3$ possible pairs which can be arranged in $2!$ ways. this gives 12 3-digit solutions.