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Each digit 1 through 5 is used exactly once to create a 5-di [#permalink]
07 Jan 2013, 05:55

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Each digit 1 through 5 is used exactly once to create a 5-digit integer. If the 3 and the 4 cannot be adjacent digits in the integer, how many 5-digit integers are possible?

Re: Each digit 1 through 5 is used exactly once to create a 5-di [#permalink]
07 Jan 2013, 09:32

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This post received KUDOS

Expert's post

daviesj wrote:

Each digit 1 through 5 is used exactly once to create a 5-digit integer. If the 3 and the 4 cannot be adjacent digits in the integer, how many 5-digit integers are possible?

(A) 48 (B) 66 (C) 72 (D) 78 (E) 90

Such questions are best solved using the complement approach. You need to find arrangements in which 3 and 4 are not adjacent. Instead, work on finding the arrangements in which they are together since it is much easier to do.

Number of arrangements using 5 distinct digits = 5! Number of arrangements in which 3 and 4 are adjacent - consider 3 and 4 together as one group. Now you have 4 numbers/groups to arrange which can be done in 4! ways. In each of these arrangements, 3 and 4 can be arranged as 34 or 43. So we need to multiply 4! by 2 to give all arrangements where 3 and 4 are adjacent to each other e.g. 34251, 43251, 23415, 32415 ... Number of arrangements in which 3 and 4 are not adjacent = 5! - 2*4! = 72

Re: Each digit 1 through 5 is used exactly once to create a 5-di [#permalink]
07 Jan 2013, 06:10

Total number of 5 digit numbers that can be formed = 5! = 120. Total number of 5 digit numbers in which 3 and 4 are adjacent = 48. This can be calculated by taking 3 and 4 as a pair and considering positions where they can appear. Number of 5 digit numbers in which 3 and 4 are not adjacent = Total number of 5 digit numbers - Total number of 5 digit numbers in which 3 and 4 are adjacent. Ans = 120 - 48 = 72.

Re: Each digit 1 through 5 is used exactly once to create a 5-di [#permalink]
14 Jan 2013, 03:46

daviesj wrote:

Each digit 1 through 5 is used exactly once to create a 5-digit integer. If the 3 and the 4 cannot be adjacent digits in the integer, how many 5-digit integers are possible?

(A) 48 (B) 66 (C) 72 (D) 78 (E) 90

Total number of arrangements of 5 digit integer = 5! = 120

Let 3 and 4 be single digit like { 1 2 X 5 } This set arrangement is 4! = 24 As 3 and 4 can be interchanged between them 2(24) = 48

So 120-48 =72 Pls correct me if my solution is wrong! _________________

GMAT - Practice, Patience, Persistence Kudos if u like

Re: Each digit 1 through 5 is used exactly once to create a 5-di [#permalink]
14 Jan 2013, 11:45

daviesj wrote:

Each digit 1 through 5 is used exactly once to create a 5-digit integer. If the 3 and the 4 cannot be adjacent digits in the integer, how many 5-digit integers are possible?

(A) 48 (B) 66 (C) 72 (D) 78 (E) 90

combinatorics and permutations eish..how do you guys make out on which way to use,whether combination or permutation,when you have such problem to solve?..when do you use permutation and when do you use combination?

Re: Each digit 1 through 5 is used exactly once to create a 5-di [#permalink]
14 Jan 2013, 19:50

Expert's post

chiccufrazer1 wrote:

daviesj wrote:

Each digit 1 through 5 is used exactly once to create a 5-digit integer. If the 3 and the 4 cannot be adjacent digits in the integer, how many 5-digit integers are possible?

(A) 48 (B) 66 (C) 72 (D) 78 (E) 90

combinatorics and permutations eish..how do you guys make out on which way to use,whether combination or permutation,when you have such problem to solve?..when do you use permutation and when do you use combination?

Re: Each digit 1 through 5 is used exactly once to create a 5-di [#permalink]
28 Nov 2014, 04:51

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