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# too many combinations?

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The product of the digits of 3214 is 24 How many distinct four-digit positive integers with unit digit not equal to 1 are there such that the product of their digits equals 18?

It seems like there are too many combinations for just brute forcing it...What's the faster method? Please explain how you got the answer.

Guest May 27, 2018
#1
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The prime factorization of 18 is $$2\cdot3^2$$

Case 1: {1, 2, 3, 3}

If these digits are used, there are 4!/2! combinations.

Case 2: {1, 1, 3, 6}

If these digits are used, there are 4!/2! combinations.

Case 3: {1, 1, 2, 9}

If these digits are used, there are 4!/2! combinations.

There are a total of 4!/2! + 4!/2! + 4!/2! = 36 positive four digit integers.

However, excluding permutations with ones in the unit digits, we get 36 - 15 = 21

As far as I know this is the best way.

I hope this helped,

Gavin

GYanggg  May 27, 2018
edited by GYanggg  May 27, 2018
#2
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1 2 3 3 =4!/2! - (3  nPr  1) = 9
1 1 3 6 =+ 4!/2! - (3  nPr  2) = 6
1 1 2 9 =+ 4!/2! - (3  nPr  2) = 6
4!/2! - (3  nPr  1) + 4!/2! - (3  nPr  2) + 4!/2! - (3  nPr  2)=21 - total permutations exluding 1 in "units" position.
(1129, 1136, 1163, 1192, 1219, 1233, 1316, 1323, 1332, 1613, 1912, 2119, 2133, 2313, 3116, 3123, 3132, 3213, 3312, 6113, 9112)=21 permutations.

Guest May 27, 2018