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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 ? (A) 48 (B) 36 (C) 72 (D) 60 (E) 120
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Re: A five digit number is to be formed using each of the digits [#permalink]
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05 Sep 2013, 22:46
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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
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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.
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Re: A five digit number is to be formed using each of the digits [#permalink]
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06 Sep 2013, 04:55
Awesome solutions MacFauz and Bunuel, +1 kudos for both. Thanks a lot



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Re: A five digit number is to be formed using each of the digits [#permalink]
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06 Sep 2013, 04:59
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Great learning on this one!! thank you very much!



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Re: A five digit number is to be formed using each of the digits [#permalink]
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06 Sep 2013, 05:07
manuelpalomares wrote: Great learning on this one!! thank you very much! No KUDOS?



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Re: A five digit number is to be formed using each of the digits [#permalink]
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Re: A five digit number is to be formed using each of the digits [#permalink]
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06 Sep 2013, 07:32
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Bunuel wrote: arakban99 wrote: manuelpalomares wrote: Great learning on this one!! thank you very much! No KUDOS? Good questions deserves kudos as much as as good solution. +1. Haha... I second that....
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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, 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=totalcases when they are together 12048=72
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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, 21:33
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. Reqd. Answer = 5!  4!2! = 120  (24*2) = 72 Answer: C



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Re: A five digit number is to be formed using each of the digits [#permalink]
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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, 22:04
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 'na' objects, and then arrange these 'a' objects in the gaps created by those 'na' 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.




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