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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

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

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