# How many $4$ digit numbers such that $4$ is always left of $1$ can be formed? (Repetition is NOT allowed)

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

• Do $1$ and $4$ have to occur? – Peter Melech Oct 28 '18 at 12:48
• Does 4 have to be directly to the left of 1, or would 4213 work? – Theo C. Oct 28 '18 at 12:49
• Doesn't $4231$ meet the criteria? You can't get this by permuting $41,3$ and $2.$ $(3$ elements, one with two digits.) – saulspatz Oct 28 '18 at 12:49
• @PeterMelech Yes, exactly. – Mark Oct 28 '18 at 12:51
• See the updated question, please. – Mark Oct 28 '18 at 12:52

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