Suppose that you are given a set
$$A = \{ 1,2,3,4,5,6\}$$
- How many $4$ digit numbers such that $4$ is always right of $1$ can be formed? (Repetition is NOT allowed)
My attempt:
$$41\_ \space \_$$
There are $3!$ ways to permutate them and ${4}\choose{2}$ ways to pick 2 numbers out of remaining $4$ numbers.
Can you assist me?
Regards
there are $\binom{4}{2}$ ways to choose the places for $4$ and $1$ and $4 \cdot 3=12$ ways to choose $2$ from the other $4$ digits and thus: $$\binom{4}{2} \cdot 12=6 \cdot 12=72$$ possible choices