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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.

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  • $\begingroup$ @abhishekchaudhary 24 looks too less..!!!! $\endgroup$
    – Kumar
    May 14, 2018 at 3:21
  • $\begingroup$ @abhishekchaudhary Can you please elaborate the solution? $\endgroup$
    – Kumar
    May 14, 2018 at 3:33
  • $\begingroup$ Your approach is correct and its the easiest try using permutations and you should get your answer $\endgroup$
    – Abhishek Choudhary
    May 14, 2018 at 3:36
  • $\begingroup$ 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 $2$. $\endgroup$
    – Robert Israel
    May 14, 2018 at 3:54
  • $\begingroup$ Also, you'll probably want to look separately at the cases where the first digit is or is not $3$. $\endgroup$
    – Robert Israel
    May 14, 2018 at 3:58

1 Answer 1

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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.

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