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A five digit number is to be formed using each of the digits [#permalink]

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05 Sep 2013, 22:29

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A five digit number is to be formed using each of the digits 1, 2, 3, 4 and 5 ONLY ONCE. How many numbers can be formed when 1 and 2 are not together ?

Re: A five digit number is to be formed using each of the digits [#permalink]

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06 Sep 2013, 00:06

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MacFauz wrote:

A five digit number is to be formed using each of the digits 1, 2, 3, 4 and 5 ONLY ONCE. How many numbers can be formed when 1 and 2 are not together ?

(A) 48 (B) 36 (C) 72 (D) 60 (E) 120

No. of ways the 5 digit number can be formed = 5! = 120

Let us fuse 1 & 2. No. of ways in which 1 and 2 are together = 4! x 2! (Multiplying by 2! because 1 and 2 can be arranged in 2! ways) = 24*2 = 48

No. of ways 1 & 2 wont be together = 120 - 48 =72

Great solution. +1.

Just to elaborate the case when 1 and 2 are together:

Glue 1 and 2 together and consider them as one unit: {12}.

Now, 4 units {12}, {3}, {4}, and {5} can be arranged in 4! ways and 1 and 2 within their unit can be arranged in 2 ways ({12} or {21}), thus total = 4!*2. _________________

Re: A five digit number is to be formed using each of the digits [#permalink]

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07 Sep 2013, 06:19

C

Just subtract from the total number of cases.

total number of combinations=5*4*3*2*1=120 When two numbers are together, we will have pack the two numbers together, therefore combinations=4!*2(numbers can be arranged themselves)

The cases where they are not together=total-cases when they are together

120-48=72 _________________

--It's one thing to get defeated, but another to accept it.

Re: A five digit number is to be formed using each of the digits [#permalink]

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10 Apr 2015, 14:18

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