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Bunuel
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The number of ways in which 5 different books can be distributed among 10 Feb 2021, 06:00
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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!


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The number of ways in which 5 different books can be distributed among 10 Feb 2021, 07:19
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Bunuel
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!
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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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The number of ways in which 5 different books can be distributed among 10 Feb 2021, 06:42
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Bunuel
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!
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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
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The number of ways in which 5 different books can be distributed among 08 Oct 2022, 22:34
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Bunuel
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

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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"if each person can get at most one book is"
Does it mean that one person may not get any books right? if yes, then we should have 11 ways for the first book to be selected.

first book 11 ways (possible to distribute to 10 persons or not at all)
second book 10 ways
third book 9 ways
forth book 8 ways
fifth book 7 ways

If I approached in this way, what is wrong?
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