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10 Feb 2021, 06:00
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
62% (01:17) correct
38%
(01:30)
wrong
based on 209
sessions
History
Date
Time
Result
Not Attempted Yet
The number of ways in which 5 different books can be distributed among 10 people if each person can get at most one book is:
(A) 252
(B) 10^5
(C) 5^10
(D) 10C5*5!
(E) 10C5*10!
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 252
(B) 10^5
(C) 5^10
(D) 10C5*5!
(E) 10C5*10!
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
10 Feb 2021, 07:19
Kudos
Bookmarks
Bunuel
Take the task of distributing the 5 books and break it into stages.
Let's let A, B, C, D, and E represent the 5 books
Stage 1: Select a person to receive book A
There are 10 people to choose from, so we can complete stage 1 in 10 ways
Stage 2: Select a person to receive book B
Since no person can receive more than 1 book, there are 9 people who can receive book B
So we can complete stage 2 in 9 ways
Stage 3: Select a person to receive book C
8 people remaining. So we can complete stage 3 in 8 ways
Following the pattern, we can complete stage 4 (selecting a person to receive book D) in 7 ways
And we can complete stage 5 (selecting a person to receive book E) in 6 ways
By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus distribute all 5 books) in (10)(9)(8)(7)(6) ways
Check the answer choices.....
Notice that D) \(10C5 \times 5!= \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times 5! \)
\(= \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times 5 \times 4 \times 3 \times 2 \times 1 \)
\(= 10 \times 9 \times 8 \times 7 \times 6 \)
Answer: D
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.
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General Discussion
10 Feb 2021, 06:42
Kudos
Bookmarks
Bunuel
each person can get at most one book
i.e. 5 books will be given to 5 different people out of 10
5 people out of 10 who get the book can be chosen in \(^{10}C_5\) ways = 252
But we need to distribute the book among them which can happen in 5! ways
So total outcomes = \(^{10}C_5*5!\)
Answer: Option D
