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How many ways can the five digits 3, 3, 4, 5, 6 be arranged into a 5-d
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21 Mar 2018, 23:01

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Question Stats:

54% (01:13) correct 46% (01:45) wrong based on 62 sessions

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How many ways can the five digits 3, 3, 4, 5, 6 be arranged into a 5-digit number so that the two occurrences of the digit 3 are separated by at least one other digit?

Re: How many ways can the five digits 3, 3, 4, 5, 6 be arranged into a 5-d
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21 Mar 2018, 23:09

Bunuel wrote:

How many ways can the five digits 3, 3, 4, 5, 6 be arranged into a 5-digit number so that the two occurrences of the digit 3 are separated by at least one other digit?

Re: How many ways can the five digits 3, 3, 4, 5, 6 be arranged into a 5-d
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23 Mar 2018, 12:44

Bunuel wrote:

How many ways can the five digits 3, 3, 4, 5, 6 be arranged into a 5-digit number so that the two occurrences of the digit 3 are separated by at least one other digit?

(A) 48 (B) 36 (C) 24 (D) 18 (E) 12

The only way that the two digits are separated by at least one other digit is if they are NOT next to each other. We can use the formula:

(Total number of ways to create the numbers) - (number of ways with the 3’s together) = number of ways to create the numbers with 3’s separated by at least one digit

Using the indistinguishable permutations formula, we note that the two 3’s are indistinguishable. Thus, the total number of ways to create the 5-digit numbers is 5!/2! = 60 ways.

Total number of ways to create the numbers with the 3’s together is 4! = 24 ways.

So, the number of ways to create the numbers with the 3’s separated by at least one digit is 60 - 24 = 36.

Answer: B
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Re: How many ways can the five digits 3, 3, 4, 5, 6 be arranged into a 5-d
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25 Mar 2018, 16:16

Top Contributor

Bunuel wrote:

How many ways can the five digits 3, 3, 4, 5, 6 be arranged into a 5-digit number so that the two occurrences of the digit 3 are separated by at least one other digit?

(A) 48 (B) 36 (C) 24 (D) 18 (E) 12

Here's another approach:

Take the task of arranging the 5 digits and break it into stages.

Stage 1: Arrange the 4, 5 and 6 We can arrange n unique objects in n! ways So, we can arrange these 3 digits in 3! ways (6 ways) So, we can complete stage 1 in 6 ways

TRICKY PART: We'll now add some spaces where the 3's can be placed. So, for example, if in stage 1, we arranged three digits as 645, then we'll add spaces before and after each digit. So, we'd get: _6_4_5_ We will place the two 3's in two of the 4 possible spaces. This will ENSURE that the 3's are not together.

Stage 2: Select two spaces in which to place the 3's Since the order in which we select the spaces does not matter, we can use combinations. We can select 2 spaces from 4 spaces in 4C2 ways (6 ways) So, we can complete stage 2 in 6 ways

By the Fundamental Counting Principle (FCP), we can complete the two stages (and thus arrange all 5 digits) in (6)(6) ways (= 36 ways)

Answer: B

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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