# 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?

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*}

• 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. Jan 23, 2022 at 22:39
• 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 $$\frac{5!}{(5-2)!}~~\text{rather than} ~~\frac{5!}{[(5-2)!] \times [2!]}.$$ Jan 23, 2022 at 23:47
• @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. Jan 23, 2022 at 23:51
• @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. Jan 23, 2022 at 23:56
• Guys , I follow your comments. If it was saying WITH replacement then I must use the 5!/((5-2!)*2!) ? Jan 23, 2022 at 23:59

## 1 Answer

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

• 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? Jan 24, 2022 at 9:00
• Because you can't have a number like 6 x x 6 , since only one $6$ is available Jan 24, 2022 at 9:04
• thank you for your answer Jan 24, 2022 at 9:05
• Glad to be of help :) Jan 24, 2022 at 10:14