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Lyndsy has six novels she wants to read, One of which is Emma, [#permalink]
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Emma should be there on the list. So start with Emma and choosing 3 out of the remaining 5 would be in 5P3 ways.

Then, arranging the list with Emma at the remaining 3 places in the list would be 5P3*4 = 240.
So option C
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Lyndsy has six novels she wants to read, One of which is Emma, [#permalink]
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LUCIFER1703 wrote:
Lyndsy has six novels she wants to read, One of which is Emma, She plans to create a reading list of four of these novels for an upcoming trip, and different orders count as different lists. How many readings list are possible if Emma has to be on the list.

A 60
B 120
C 240
D 360
E 6

Please hit a Kudos if you like the post.


Out of 4 novels,Emma is already selected.Hence we need to select 3 out of the remaining 5 novels in 5C3 ways..Now the total 4 novels can also be arranged in 4! ways..No of arrangements will be 5c3* 4!=10*24=240 ways

Originally posted by ahamd on 08 Mar 2023, 08:32.
Last edited by ahamd on 11 Mar 2023, 07:21, edited 1 time in total.
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Re: Lyndsy has six novels she wants to read, One of which is Emma, [#permalink]
Different than above given ways to solve thus:

1. Different ways to create reading list of 4 books out of 6 = 6x5x4x3 = 360 ways

2. Different ways to create a reading list of 4 books out of 5 books (excluding Emma)
5x4x3x2 =120 ways

Different ways in which Emma is on the list of 4 books

(1) - (2) = 360-120 =240

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Re: Lyndsy has six novels she wants to read, One of which is Emma, [#permalink]
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