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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, 21:29

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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 ?

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

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07 Sep 2013, 05:19

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

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10 Apr 2015, 13:18

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

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30 May 2016, 20:09

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

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30 May 2016, 20:33

arakban99 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

Reqd ways = (Total No. of ways - cases when both 1 and 2 are together)

5 numbers can be arranged in 5 places in 5! ways. Now tie 1 and 2 together so effectively there are 4 nos. they can be arranged in 4! ways. 1 and 2 can be arranged within themselves in 2!ways.

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

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21 Jun 2017, 14:46

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

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21 Jun 2017, 21:04

Another way to ensure that 1 and 2 are not together is to use 'Gaps' method. Basically, when we are given 'n' objects to be arranged so that 'a' out of them should NOT be together - we first arrange the remaining 'n-a' objects, and then arrange these 'a' objects in the gaps created by those 'n-a' objects.

I will try to explain by this question.

We have to arrange 1, 2, 3, 4, 5 so that 1 and 2 are not together. So lets first arrange the remaining numbers. Remaining 3 numbers can be arranged in 3! = 6 ways.

Now, for each of these 6 ways, the arranged numbers create 4 gaps: ...3...4...5...

one gap before 3, another gap between 3 and 4, another gap between 4 and 5, and the last gap after 5.

So for every arrangement of those 3 numbers, we have 4 gaps - and we can now place the numbers '1' and '2' in any two out of these 4 gaps (and thus they will not touch each other). Number of ways of doing that = 4P2 (arranging 4 objects at 2 places) = 12 ways.

Thus total ways = 3! * 4P2 = 6*12 = 72. Hence C answer

PS - this same can be extended to any such question. So if we have 7 people out of which A,B,C have to be separated (no two of them can sit together), we can first arrange remaining 4 people in 4! ways, and now out of the 5 gaps created (4 people in a line will create 5 gaps), we can place A,B,C in any of the 3 gaps out of 5 in 5P3 ways (60). Thus total ways = 4! * 5P3 = 24 * 60 = 1440.

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

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23 Sep 2017, 21:57

arakban99 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 ?