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Five letters A, P, P, L and E are listed in a row. How many

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by Max@Math Revolution » Sun Sep 30, 2018 11:36 pm
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[Math Revolution GMAT math practice question]

Five letters A, P, P, L and E are listed in a row. How many arrangements have at least one letter between the two Ps?

A. 24
B. 30
C. 36
D. 42
E. 48
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by GMATGuruNY » Mon Oct 01, 2018 5:03 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Five letters A, P, P, L and E are listed in a row. How many arrangements have at least one letter between the two Ps?

A. 24
B. 30
C. 36
D. 42
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 P's:
5!/2! = 60.

Bad arrangements:
In a bad arrangement, the two P's are in adjacent slots.
Let [PP] represent the 2 adjacent P's.
Number of ways to arrange the 4 elements [PP], D, G and T = 4! = 24.

Good arrangements:
Total arrangements - bad arrangements = 60-24 = 36.

The correct answer is C.
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Five letters A, P, P, L and E are listed in a row. How many

by Brent@GMATPrepNow » Mon Oct 01, 2018 8:53 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Five letters A, P, P, L and E are listed in a row. How many arrangements have at least one letter between the two Ps?

A. 24
B. 30
C. 36
D. 42
E. 48
Here's an approach that doesn't require us to subtract the bad arrangements.

Take the task of arranging the 5 letters and break it into stages.

Stage 1: Arrange the letters A, L, E in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 3 letters in 3! ways (= 6 ways)
So, we can complete stage 1 in 6 ways

IMPORTANT: For each arrangement of 3 letters (above), there are 4 places where the two P's can be placed.
For example, in the arrangement AEL, we can add spaces as follows _A_E_L_
So, if we place each P in one of the available spaces, we can ENSURE that the two P's are never together.

Stage 2: Select two available spaces and place an P 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: C

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: https://www.gmatprepnow.com/module/gmat- ... /video/775

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

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should ... 67256.html
- https://www.beatthegmat.com/counting-pr ... 44302.html
- https://www.beatthegmat.com/picking-a-5 ... 73110.html
- https://www.beatthegmat.com/permutation ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatori ... 73194.html
- https://www.beatthegmat.com/arabian-hor ... 50703.html
- https://www.beatthegmat.com/sub-sets-pr ... 73337.html
- https://www.beatthegmat.com/combinatori ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-se ... 71047.html
- https://www.beatthegmat.com/combinatori ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p ... 71001.html
- https://www.beatthegmat.com/permutation ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladi ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by fskilnik@GMATH » Mon Oct 01, 2018 12:11 pm
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Five letters A, P, P, L and E are listed in a row. How many arrangements have at least one letter between the two Ps?

A. 24
B. 30
C. 36
D. 42
E. 48
\[?\,\,\,:\,\,\,\# \,\,P{\text{s}}\,\,{\text{not}}\,\,{\text{together}}\]
Total number of permutations of 5 elements, with exactly one repeating element appearing twice:
\[\frac{{5!}}{{2!}} = 5 \cdot 4 \cdot 3\]
Total number of permutations of 3 "single elements" (A,L,E) and one "double element" (PP):
\[4!\]
(We took into account the two P´s are indistinguishable, of course!)

\[? = 5 \cdot 4 \cdot 3 - 4 \cdot 3 \cdot 2 = 12\left( {5 - 2} \right) = 36\]

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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by Max@Math Revolution » Wed Oct 03, 2018 1:10 am
=>

When we encounter “at least” in counting questions or probability questions, we should consider complementary counting.
The total number of arrangements of the 5 letters is 5!/2! (5! Counts each arrangement of the two Ps 2! times).
The number of arrangements with no letter between the two Ps is 4!.
Thus, the number of arrangements in which at least one letter lies between the two Ps is 5!/2! – 4! = 60 – 24 = 36.

Therefore, C is the answer.
Answer: C
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