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How Many Ways can G,H,I,I, and J be arranged ?

Expert replies
by nitya34 » Fri Jun 26, 2009 6:33 pm
in how many ways can G,H,I,I, and J be arranged to compose a 5-letter code so that two "I" are not next to each other?


OA[spoiler]=60-24=36[/spoiler]
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Source: — Problem Solving |

by DanaJ » Sat Jun 27, 2009 12:30 am
Total ways of arranging the 5 letters will be 5!/2! = 60.
Then you need to subtract the cases where you have the two I's next to each other. In such cases, they'd create one single, indivisible group. So you can safely say that the number of cases in which you have II is the number of possible arrangements of (G), (H), (II) and (J) or possible arrangements of 4 objects, i.e. 4! = 24.
You're left with the answer you provided.
Last edited by DanaJ on Sat Jun 27, 2009 1:50 am, edited 1 time in total.
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by rahulg83 » Sat Jun 27, 2009 1:11 am
DanaJ wrote:Total ways of arranging the 5 letters will be 5! = 60.
Then you need to subtract the cases where you have the two I's next to each other. In such cases, they'd create one single, indivisible group. So you can safely say that the number of cases in which you have II is the number of possible arrangements of (G), (H), (II) and (J) or possible arrangements of 4 objects, i.e. 4! = 24.
You're left with the answer you provided.
Well total ways of arranging 5 letters with 2 common letters=5!/2!=60 and 4!=24
Answer should be 36
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by DanaJ » Sat Jun 27, 2009 1:49 am
Yeah, you're right. I was thinking it like that, but didn't exactly write it down as such.
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by yogami » Sat Jun 27, 2009 9:17 am
Hey guys why are you dividing 5!/2!?
I thought the answer was 5! - 4! = 96. Why will you divide by 2! when the letters are common??
I simply took it as. Since they are five letters, the total number of ways they can be arranged with repetitions are 5! (without repetitions they will be 5!/2 though) so with repetitions the answer is 96 and without it is yeah 36
200 or 800. It don't matter no more.
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by shibal » Sat Jun 27, 2009 2:52 pm
why the 5 ways to arrange 5 letters 60? shouldn't be 5! only?
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by DanaJ » Sun Jun 28, 2009 4:22 am
You need to divide by 2! because you have two letters that are basically the same: I and I. Say you have I(1) and I(2). There's no difference between HI(1)I(2)GJ ans HI(2)I(1)GJ, since you won't see those numbers when it comes down to it.
5! applies only when you do not have any repeated letters.
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