Og 2017 problem
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Rather than cite the question number, please post the problem itself, along with the answer choices.
Here is the problem:
Total arrangements:
Number of ways to arrange 5 elements = 5!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical I's:
5!/2! = 60.
Bad arrangements:
In a bad arrangement, the two I's are in adjacent slots.
Let [II] represent the 2 adjacent I's.
Number of ways to arrange the 4 elements [II], D, G and T = 4! = 24.
Good arrangements:
Total arrangements - bad arrangements = 60-24 = 36.
The correct answer is D.
Here is the problem:
Good arrangements = total arrangements - bad arrangements.The letters D, G, I, I, and T can be used to form 5-letter strings such as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
(A) 12
(B) 18
(C) 24
(D) 36
(E) 48
Total arrangements:
Number of ways to arrange 5 elements = 5!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical I's:
5!/2! = 60.
Bad arrangements:
In a bad arrangement, the two I's are in adjacent slots.
Let [II] represent the 2 adjacent I's.
Number of ways to arrange the 4 elements [II], D, G and T = 4! = 24.
Good arrangements:
Total arrangements - bad arrangements = 60-24 = 36.
The correct answer is D.
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Hi Vishnu1994,
Since the two "I"s cannot be side-by-side, there are a limited number of ways to arrange the 5 letters. As such, with a little permutation math and some 'brute force', we can map out the possibilities:
_ _ _ _ _
If the first letter is an I, then the second letter CANNOT be an I (it would have to be one of the other 3 non-I letters)...
i 3
From here, any of the remaining letters can be in the 3rd spot. After placing one, either of the remaining two letters can be in the 4th spot and the last letter would be in the 5th spot...
i 3 3 2 1
This would give us (3)(3)(2)(1) = 18 possible arrangements with an I in the 1st spot.
If a non-I is in the 1st spot and an I is in the 2nd spot, then we have...
3 i _ _ _
A non-I would have to be in the 3rd spot, then either remaining letter could be 4th...
3 i 2 2 1
This would give us (3)(2)(2)(1) = 12 possible arrangements
Next, we could have two non-Is to start off, then Is in the 3rd and 5th spots...
3 2 i 1 i
This would give us (3)(2)(1) = 6 possible arrangements
There are no other options to account for, so we have 18+12+6 total arrangements.
Final Answer: D
GMAT assassins aren't born, they're made,
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Since the two "I"s cannot be side-by-side, there are a limited number of ways to arrange the 5 letters. As such, with a little permutation math and some 'brute force', we can map out the possibilities:
_ _ _ _ _
If the first letter is an I, then the second letter CANNOT be an I (it would have to be one of the other 3 non-I letters)...
i 3
From here, any of the remaining letters can be in the 3rd spot. After placing one, either of the remaining two letters can be in the 4th spot and the last letter would be in the 5th spot...
i 3 3 2 1
This would give us (3)(3)(2)(1) = 18 possible arrangements with an I in the 1st spot.
If a non-I is in the 1st spot and an I is in the 2nd spot, then we have...
3 i _ _ _
A non-I would have to be in the 3rd spot, then either remaining letter could be 4th...
3 i 2 2 1
This would give us (3)(2)(2)(1) = 12 possible arrangements
Next, we could have two non-Is to start off, then Is in the 3rd and 5th spots...
3 2 i 1 i
This would give us (3)(2)(1) = 6 possible arrangements
There are no other options to account for, so we have 18+12+6 total arrangements.
Final Answer: D
GMAT assassins aren't born, they're made,
Rich