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arragements

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by figs » Tue Apr 07, 2009 10:18 am
How many different possible arrangements can be obtained from the letters G, M, A, T, I, I, and T, such that there is at least one character between both I’s?

A. 360
B. 720
C. 900
D. 1,800
E. 5,040
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by scoobydooby » Tue Apr 07, 2009 10:27 am
atleast 1 character between Is=> Is should not be together.

we shall find out total arrangements and deduct cases where Is are together to get cases where 1s are not together.

total arrangements possible: 7!/(2!*2!)=1260 ways.
(there are 7 letters in all, 2 letters repeated twice.)

no of arrangements in which Is are together: consider the Is as a unit so there are in effect 6 letters with Ts repeated twice
so 6!/2!=360 ways.

no of arrangments where Is are not together: 1260-360= 900 ways.

C?
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by figs » Tue Apr 07, 2009 11:44 am
OA: C
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by pakaskwa » Tue Apr 07, 2009 12:39 pm
Step 1: calculate all possible ways to arrange 7 letters including 2 same letters.
All possible ways to arrange 7 DIFFERENT letters is:
7!=5040

But, remember there are 2 same letters I. So 7! contains duplicate arrangements. For example,
GMATTII is the same as GMATTII (where two Is were switched position)

Therefore, all possible ways to arrange 7 letters including 2 same letters is:
7!/2=2520

Step 2: calculate all possible ways to arrange 6 different letters (when two Is are never seperated, they can be considered as ONE letter)
6!=720

Step 3: 2520-720=1800
This is all possible arrangements when two Is are separated.

Answer is D.
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