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Permutations Question

This topic has 5 expert replies and 3 member replies
It'sGMATtime Newbie | Next Rank: 10 Posts Default Avatar
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Permutations Question

Post Tue Jun 28, 2016 11:10 am
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

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GMAT/MBA Expert

Post Sat Jul 02, 2016 9:20 am
Hi needgmat,

When doing a permutation, you have to make sure that you don't count duplicate results. Here's a simple example:

IF we had 3 different letters (A, B, C) we could arrange them in 3! = 6 different ways:

ABC
ACB
BAC
BCA
CAB
CBA

IF we had a pair of duplicate letters though (A, A, B), then there would NOT be 6 different arrangements:

AAB
ABA
BAA

There's no different between the "first A" and the "second A", so we can't count an option twice (AAB and AAB are the same result). We have to divide the 3! by 2! (for the 2 As). 3!/2! = 3 options.

The same idea exists behind this question. We have 5 total letters but a pair of Is:

5!/2! = 60 arrangements

GMAT assassins aren't born, they're made,
Rich

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GMAT/MBA Expert

Post Wed Aug 23, 2017 4:45 pm
It'sGMATtime wrote:
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
Take the task of arranging the 5 letters and break it into stages.

Stage 1: Arrange the 3 CONSONANTS (D, G and T) in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 3 consonants in 3! ways (= 6 ways)
So, we can complete stage 1 in 6 ways

IMPORTANT: For each arrangement of 3 consonants, there are 4 places where the two I's can be placed.
For example, in the arrangement DTG, we can add spaces as follows _D_T_G_
So, if we place each I in one of the available spaces, we can ENSURE that the two I's are never together.

Stage 2: Select two available spaces and place an I in each space.
Since the order in which we select the two spaces does not matter, we can use combinations.
We can select 2 spaces from 4 spaces in 4C2 ways (= 6 ways)
So we can complete stage 2 in 6 ways.

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange all 5 letters) in (6)(6) ways (= 36 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting/video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat-counting/video/776

Then you can try solving the following questions:

EASY
- http://www.beatthegmat.com/what-should-be-the-answer-t267256.html
- http://www.beatthegmat.com/counting-problem-company-recruitment-t244302.html
- http://www.beatthegmat.com/picking-a-5-digit-code-with-an-odd-middle-digit-t273110.html
- http://www.beatthegmat.com/permutation-combination-simple-one-t257412.html
- http://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- http://www.beatthegmat.com/combinatorics-solution-explanation-t273194.html
- http://www.beatthegmat.com/arabian-horses-good-one-t150703.html
- http://www.beatthegmat.com/sub-sets-probability-t273337.html
- http://www.beatthegmat.com/combinatorics-problem-t273180.html
- http://www.beatthegmat.com/digits-numbers-t270127.html
- http://www.beatthegmat.com/doubt-on-separator-method-t271047.html
- http://www.beatthegmat.com/combinatorics-problem-t267079.html


DIFFICULT
- http://www.beatthegmat.com/wonderful-p-c-ques-t271001.html
- http://www.beatthegmat.com/permutation-and-combination-t273915.html
- http://www.beatthegmat.com/permutation-t122873.html
- http://www.beatthegmat.com/no-two-ladies-sit-together-t275661.html
- http://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

_________________
Brent Hanneson – Founder of GMATPrepNow.com
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GMAT/MBA Expert

Post Fri Dec 08, 2017 6:52 am
It'sGMATtime wrote:
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
This is a permutation problem, because the order of the letters matters. Let’s first determine how many ways we can arrange the letters. Since there are 2 repeating I’s, we can arrange the letters in 5!/2! = 120/2 = 60 ways.

We also have the following equation:

60 = (number of ways to arrange the letters with the I’s together) + (number of ways without the I’s together).

Let’s determine the number of ways to arrange the letters with the I’s together.

We have: [I-I] [D] [G] [T]

We see that with the I’s together, we have 4! = 24 ways to arrange the letters.

Thus, the number of ways to arrange the letters without the I’s together (i.e., with the I’s separated) is 60 - 24 = 36.

Answer:D

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Needgmat Master | Next Rank: 500 Posts Default Avatar
Joined
31 May 2016
Posted:
234 messages
Upvotes:
3
Post Sat Jul 02, 2016 12:06 am
Quote:
Good arrangements = total arrangements - bad arrangements.

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.
Hi GMATGuruNY ,

Can you please explain more why have we to do 5!/2!? I didn't get that part.

Many thanks in advance.

Kavin

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800_or_bust Master | Next Rank: 500 Posts Default Avatar
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Post Wed Jun 29, 2016 6:32 am
It'sGMATtime wrote:
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
An alternative approach using probabilities:

Total number of possible arrangements: 5!/2! = 120/2 = 60.

Probability that both I's are right next to each other: (2/5)(1/4) + (3/5)(1/2) = 2/5.

Probability that both I's are separated by at least one other letter: 1 - 2/5 = 3/5.

Total number of cases in which the I's are separated by at least one letter: 60 x 3/5 = 36.

Answer: D

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Post Tue Jun 28, 2016 9:42 pm
Hi It'sGMATtime,

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

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Contact Rich at Rich.C@empowergmat.com

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Post Tue Jun 28, 2016 11:49 am
It'sGMATtime wrote:
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
Good arrangements = total arrangements - bad arrangements.

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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Danny@GMATAcademy Junior | Next Rank: 30 Posts Default Avatar
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Post Tue Jun 28, 2016 11:16 am
A couple of video explanations for this question:

"Indirect" Method: https://www.youtube.com/watch?v=5LDYpUWG2HY

"Direct" Method: https://www.youtube.com/watch?v=TSpp5XbJE3A

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Last edited by Danny@GMATAcademy on Sat Jan 07, 2017 6:58 am; edited 1 time in total

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