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I was solving some exercises and I would like to hear your opinion about this.

Question 1 : We are given the digits $1,2,4,5,6,7,9$. How many even, $4$-digit numbers, bigger than $5000$, with distinct digits can we form?

What I did :

Step 1: It is Combinations nCr, with formula $nCr = C(n,r) = \frac{n!}{r!(n - r)!}$. The order doesn't matter and replacement not allowed-cause it says needs only 1 time to be used each digit.

Step 2: I have $7$ digits ($1,2,4,5,6,7,9$). I have to chose $4$ digits so the $r=4$.

Step 3: How to create even numbers? the digit $4$ with digit $2$ if I add them (+) it makes $6$, (I can't use the digit $4$ and $2$). The digit $7 + 1$ makes $8$, (so those digits can't be used again). Now I add $5 + 6 + 9= 20$, so I have $6820$, my $4$ digits.

Step 4: I check if $5000 < 6820$. It is true.

Step 5: Now I am not sure if $n=6820$? and $r=4$. If that is true, \begin{align*} C(n,r) & = C(6820,4)\\ & = \frac{6820!}{4!(6820−4)!}\\ & = \frac{6820!}{4!6816!} \end{align*}

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    $\begingroup$ Picking an even number means the units digit must be an even number. In this case, it must be $2, 4$, or $6$. The importance of the fact that the number must be at least $5000$ is that if you pick $2$ or $4$, you still have four choices for the thousands digit. On the other hand, if you pick $6$, you have three choices for the thousands digit since you can no longer choose $6$. This problem required the use of the Multiplication Principle and the Addition Principle, nothing more. $\endgroup$ Jan 23, 2022 at 22:39
  • $\begingroup$ The comment of @N.F.Taussig nailed it. Unsure, from your posting, whether you already also realize the following: The other thing to consider is that in general, you are interested in permutations, not combinations. For example, consider the simpler question of how many $3$ digit numbers can be constructed by choosing $3$ numbers, without replacement, from the set $\{1,2,3,4,5\}.$ The answer is $$\frac{5!}{(5-2)!}~~\text{rather than} ~~\frac{5!}{[(5-2)!] \times [2!]}.$$ $\endgroup$ Jan 23, 2022 at 23:47
  • $\begingroup$ @user2661923 In this case, the constraints that the number is even and larger than $5000$ means we have to use the Multiplication and Addition Principles, not permutations. $\endgroup$ Jan 23, 2022 at 23:51
  • $\begingroup$ @N.F.Taussig I agree. However, it is unclear whether the OP (i.e. original poster) understands this. Permutations may be considered to be an application of the Multiplication principle, where there are no constraints. I did think that it was important to emphasize that in the simpler problem that I posed, the enumeration is $(20)$, rather than $(10)$. It is unclear whether the OP realizes this. $\endgroup$ Jan 23, 2022 at 23:56
  • $\begingroup$ Guys , I follow your comments. If it was saying WITH replacement then I must use the 5!/((5-2!)*2!) ? $\endgroup$
    – ek.Sek
    Jan 23, 2022 at 23:59

1 Answer 1

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A neat way to do it is to realize that

  • Whichever way you choose other digits, there will be $(5\cdot4)=20$ ways to fill the two middle digits

  • You can't have $6$ at both the thousands and units place, so ways of filling these digits would be $[(4\cdot3)-1] = 11$

Multiply the two

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  • $\begingroup$ Multiply the two,means 20ways of fill the two middle digits and 11 I found the other so it would be 20*11 right?also why you -1 from here (4⋅3)−1]=11? $\endgroup$
    – ek.Sek
    Jan 24, 2022 at 9:00
  • $\begingroup$ Because you can't have a number like 6 x x 6 , since only one $6$ is available $\endgroup$ Jan 24, 2022 at 9:04
  • $\begingroup$ thank you for your answer $\endgroup$
    – ek.Sek
    Jan 24, 2022 at 9:05
  • $\begingroup$ Glad to be of help :) $\endgroup$ Jan 24, 2022 at 10:14

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