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How many different positive integers having six digits are

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by BTGmoderatorDC » Mon Aug 27, 2018 6:14 pm

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How many different positive integers having six digits are there, where exactly on of the digits is a 3, exactly two of the digits are a 4, exactly one of the digits is a 5, and each of the other digits is a 7 or an 8?

A) 360
B) 720
C) 840
D) 1,080
E) 1,440

OA B

Source: Princeton Review
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by GMATGuruNY » Mon Aug 27, 2018 6:39 pm
BTGmoderatorDC wrote:How many different positive integers having six digits are there, where exactly on of the digits is a 3, exactly two of the digits are a 4, exactly one of the digits is a 5, and each of the other digits is a 7 or an 8?

A) 360
B) 720
C) 840
D) 1,080
E) 1,440
There are 6 positions in the integer.

The two 4's must occupy a pair of positions.
From the 6 positions, the number of ways to choose a pair for the two 4's = 6C2 = (6*5)/(2*1) = 15.

The remaining 4 positions must be occupied by 3, 5, 7 or 8, 7 or 8.

Number of options for the 3 = 4. (Any of the 4 remaining positions.)
Number of options for the 5 = 3. (Any of the 3 remaining positions.)
Number of options for the next position = 2. (Must be 7 or 8.)
Number of options for the last position = 2. (Must be 7 or 8.)

To combine the options above, we multiply:
15*4*3*2*2 = 720.

The correct answer is B.
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by GMATGuruNY » Tue Aug 28, 2018 2:19 am
How many ways can the letters BBBGG be arranged?

Number of ways to arrange 5 elements = 5!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 3! to account for the three identical B's and by 2! to account for the two identical G's:
5!/(3!2!) = 10.
BTGmoderatorDC wrote:How many different positive integers having six digits are there, where exactly on of the digits is a 3, exactly two of the digits are a 4, exactly one of the digits is a 5, and each of the other digits is a 7 or an 8?

A) 360
B) 720
C) 840
D) 1,080
E) 1,440
Case 1: The digits are 4, 4, 3, 5, 7, 7
Since the 6-digit arrangement includes two identical 4's and two identical 7's, the number of possible arrangements = 6!/(2!2!) = 180.

Case 2: The digits are 4, 4, 3, 5, 8, 8
Since the 6-digit arrangement includes two identical 4's and two identical 8's, the number of possible arrangements = 6!/(2!2!) = 180.

Case 3: The digits are 4, 4, 3, 5, 7, 8
Since the 6-digit arrangement includes two identical 4's, the number of possible arrangements = 6!/2! = 360.

Total integers = Case 1 + Case 2 + Case 3 = 180+180+360 = 720.

The correct answer is B.
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by Scott@TargetTestPrep » Thu Apr 18, 2019 5:42 pm
BTGmoderatorDC wrote:How many different positive integers having six digits are there, where exactly on of the digits is a 3, exactly two of the digits are a 4, exactly one of the digits is a 5, and each of the other digits is a 7 or an 8?

A) 360
B) 720
C) 840
D) 1,080
E) 1,440

OA B

Source: Princeton Review
We need to determine the number of 6-digit numbers in which one of the digits is a 3, exactly two of the digits are a 4, exactly one of the digits is a 5, and each of the other digits is a 7 or an 8.

Option #1: The number has two 7s

3-4-4-5-7-7

We use the indistinguishable permutations formula to determine the number of ways to arrange 3-4-4-5-7-7:

6!/(2! x 2!) = (6 x 5 x 4 x 3 x 2)/(2 x 2) = 6 x 5 x 3 x 2 = 180

Option #2: The number has two 8s

3-4-4-5-8-8

6!/(2! x 2!) = 180

Option #3: The number has a 7 and an 8

3-4-4-5-7-8

6!/2! = (6 x 5 x 4 x 3 x 2)/2 = 6 x 5 x 4 x 3 = 360

Thus, the number can be created in 180 + 180 + 360 = 720 ways.

Answer: B

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