positive integers

GmatKiss wrote:How many positive integers less than 10,000 are such that the product of their digits is 210?

(A) 24
(B) 30
(C) 48
(D) 54
(E) 72

Edited:

210 = 2x3x5x7

We need to consider 3 cases:

Case 1: 4-digit numbers using 2, 3, 5, 7
There are 4 digits, so this can be accomplished in 4! (24) ways

Aside: Notice that 2x3=6

Case 2: 4-digit numbers using 1, 6, 5, 7
There are 4 digits, so this can be accomplished in 4! (24) ways

Case 3: 3-digit numbers using 6, 5, 7
There are 3 digits, so this can be accomplished in 3! (6) ways

Add up all 3 cases to get 24 + 24 + 6 = 54

So, the answer is D

Cheers,
Brent

Edited:

I should point out that, in the above solution, I used the nice rule that says "n unique objects can be arranged in n! ways"

Alternatively, if we don't know the above rule, we can use the Fundamental Counting Principle (FCP). For example, in case 1, we need to arrange the digits 2, 3, 5, 7.
To apply the FRC, we'll take the task of arranging the 4 digits and break it into 4 stages:

Stage 1: Select a digit for the units position.
Stage 2: Select a digit for the tens position.
Stage 3: Select a digit for the hundreds position.
Stage 4: Select a digit for the thousands position.

Stage 1: there are 4 digits, so this stage can be accomplished in 4 ways
Stage 2: once we have completed stage 1, there are 3 digits remaining, so stage 2 can be accomplished in 3 ways
Stage 3: once we have completed stage 2, there are 2 digits remaining, so stage 3 can be accomplished in 2 ways
Stage 4: once we have completed stage 3, there is 1 digit remaining, so stage 4 can be accomplished in 1 ways

From here, the FCP tells us that all 4 stages can be completed in 4x3x2x1 ways (24 ways)

If you're interested in learning more, here's a free video on the FCP: https://www.gmatprepnow.com/module/gmat-counting?id=775

Cheers,
Brent

GMATGuruNY wrote:

GmatKiss wrote:How many positive integers less than 10,000 are such that the product of their digits is 210?

(A) 24
(B) 30
(C) 48
(D) 54
(E) 72

210 = 2*3*5*7.

Case 1: 4-digit integers composed of the digits 2, 3, 5 and 7.
Number of ways to arrange these 4 digits = 4! = 24.

Case 2: 4-digit integers composed of the digits 1, 5, 6, and 7.
Number of ways to arrange these 4 digits = 4! = 24.

Case 3: 3-digit integers composed of the digits 5, 6 and 7.
Number of ways to arrange these 3 digits = 3! = 6.

Total ways = 24+24+6 = 54.

The correct answer is D.

Good catch - I completely missed the fact that 2x3 creates a new digit. My bad

Cheers,
Brent