In how many ways can the letters D, I, G, I, T be arranged

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swerve: Legendary Member; Posts: 2545; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Thu Dec 20, 2018 9:17 am

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In how many ways can the letters D, I, G, I, T be arranged so that the two I's are not next to each other?

A. 36
B. 48
C. 72
D. 96
E. 128

The OA is A

Source: Economist GMAT

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Source: Beat The GMAT — Problem Solving | Thu Dec 20, 2018 9:17 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu Dec 20, 2018 11:50 am

swerve wrote:In how many ways can the letters D, I, G, I, T be arranged so that the two I's are not next to each other?

A. 36
B. 48
C. 72
D. 96
E. 128

The OA is A

Source: Economist GMAT

One approach is to count all possible arrangements (that may or may not break the given rule) and then subtract all outcomes the break the rule.

Here's another approach.

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: A

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Fri Dec 21, 2018 4:05 pm

swerve wrote:In how many ways can the letters D, I, G, I, T be arranged so that the two I's are not next to each other?

A. 36
B. 48
C. 72
D. 96
E. 128
Source: Economist GMAT

$$?\,\,\,:\,\,\,{\rm{\# }}\,\,\,{\rm{permutations}}\,\,{\rm{Is}}\,\,{\rm{apart}}$$
$${\rm{all}}\,\,{\rm{permutations}}\,\,{\rm{with}}\,\,{\rm{repetition}} = {{5!} \over {2!}} = 5 \cdot 4 \cdot 3 = 60$$
$${\rm{all}}\,\,{\rm{permutations}}\,\,{\rm{with}}\,\,{\rm{Is}}\,\,{\rm{together}} = 4! = 24$$
$$? = 60 - 24 = 36$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8152; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Sat Mar 02, 2019 8:07 am

swerve wrote:In how many ways can the letters D, I, G, I, T be arranged so that the two I's are not next to each other?

A. 36
B. 48
C. 72
D. 96
E. 128

The OA is A

Source: Economist GMAT

We can use the equation:

Number of ways the letters D, I, G, I, T be arranged so that the two I's are not next to each other = total number of ways to arrange the letters - number of ways the two I's are next to each other.

The total number of ways to arrange the letters is:

5!/2! = 120/2 = 60

The number of ways the two I's are next to each other is:

Our arrangement looks like the following:

[I-I]-D-G-T

The number of ways to arrange [I-I]-D-G-T is 4! = 24.

So the total number of ways to arrange the letters with the I's not next to each other is 60 - 24 = 36.

Answer: A

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members; Location: Palo Alto, CA; Thanked: 2867 times; GMAT Score:800

by [email protected] » Sat Mar 02, 2019 1:07 pm

Hi All,

We're asked to find the number of ways that the letters D, I, G, I, T be arranged so that the two I's are NOT next to each other. 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: A

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

Contact Rich at [email protected]

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