Counting number of 11 digit numbers

How many even 11-digit numbers (no leading zeros) have at least three 3's?

First I tried to find the total number of 11-digit even numbers, then I have subtracted number of 11-digit numbers which have at most two 3's.

But this process looks lengthy. If there is an alternative, let me know.

• @abhishekchaudhary 24 looks too less..!!!! May 14, 2018 at 3:21
• @abhishekchaudhary Can you please elaborate the solution? May 14, 2018 at 3:33
• Your approach is correct and its the easiest try using permutations and you should get your answer May 14, 2018 at 3:36
• I hope you subtracted the number of $11$-digit even numbers with at most two $3$'s. Note that "at most two" is $0$, $1$ or $2$. May 14, 2018 at 3:54
• Also, you'll probably want to look separately at the cases where the first digit is or is not $3$. May 14, 2018 at 3:58

Your idea is a good one and is the one that I would recommend.

Let us look at the sub-problem of counting how many even 11-digit have exactly two $3$'s:

Break into cases based on whether the leading digit is a $3$ or if it is a non-$3$

• Case 1: The leading digit is a $3$

• Set the first digit as a $3$ (One option)
• Choose the final digit, it must be even (five options)
• Choose the location of the remaining $3$ (nine options)
• From left-to-right, choose the value of each remaining non-3 digit (nine options each)
• Case 2: The leading digit is not a $3$

• Choose the first digit remembering it can't be $3$ and it can't be $0$ (eight options)
• Choose the final digit, it must be even (five options)
• Choose the location of the two threes (nine choose two options)
• From left-to-right, choose the value of each remaining non-3 non-leading digit (nine options each)

$5\cdot 9^{9}+8\cdot 5\cdot \binom{9}{2}\cdot 9^7$

Apply multiplication principle and addition principle to arrive at the total number of even 11-digit numbers containing exactly two threes. Do so similarly for the other subproblems of having exactly one three or exactly zero threes.