Romi has a collection of 10 distinct books out of which 8 are small an

Question

Romi has a collection of 10 distinct books out of which 8 are small and 2 are large. In how many ways can he select 5 books to take with him on a trip if he can take at most 1 large book?

A. 56 
 B. 126 
 C. 152 
 D. 196 
 E. 252

Bunuel · Accepted Answer

Official Solution:

Romi has a collection of 10 distinct books out of which 8 are small and 2 are large. In how many ways can he select 5 books to take with him on a trip if he can take at most 1 large book?

A. 56
B. 126
C. 152
D. 196
E. 252

Since Romi can take at most 1 large book, then he can take with him either 5 small books OR 4 small books and 1 large book:  C^5_8+C^4_8*C^1_2=56+140=196 .

Answer: D

fluke · Answer

Sol:

In how many ways can Romi select 5 books:

5 Small books out of 8 AND 0 large book out of 2 books
OR
4 Small book out of 8 AND 1 large book out of 2 books

We know AND represents multiplication and OR addition;

Thus; to convert the above statement into combinations:

C^8_5*C^2_0 + C^8_4*C^2_1

\frac{8!}{5!3!}*\frac{2!}{2!0!} + \frac{8!}{4!4!}*\frac{2!}{1!1!}

56 + 140 = 196

Ans: "D"

*************************

Sorry, I have no idea about Slot Method!!!

miguelmick · Answer

If he is to select 4 small books and 1 large book, then the number of arrangements is: 8! / 4!4! * 2!/1!1!, where the first term is the number of ways in which he can pick the 4 small books and the second term is the number of ways in which he can pick the large books. This results 140. However, he can also pick 5 small books. The number of arrangements in which he can pick the 5 books will be 8!/5!3!, which results in 8*7 = 56.

Now, you have to sum both possibilities, which yields 196.

Hope it helped.

Now, you have to sum both possibilities, which yields 196.

Hope it helped.

anshumishra · Answer

Here is how you understand the given solution.
We just want to make sure that out of the 2 larger books , we don't select both of them (we can select 0 or 1 but not both).
So the number of ways = select any 5 books - select both the larger books and any 3 smaller books.

Hope that explains your question.

Posted from my mobile device

TwoThrones · Answer

Ladies and gentlemen! I highly appreciate your help. It seems using the Manhattan GMAT Word Translations strategy was too tricky for me. I used the nCr approach instead of the slot method (I had to do a little research because I hadn't done this since high school) and found out how much easier the question was. Thank you very much everyone!

gmat1220 · Answer

Yeah first select 5 books i.e. 10C5 and then subtract the remainder i.e. All big + 3 small 
10C5 - 8C3 = 196

Gladiator59 · Answer

Replying to a PM :

On first sight, you might fall for the trap that there is only one possibility - 4 small books and 1 large book. However, given information that Romi has space for only one large book (and assuming that one small book can replace one large book by common sense), there are actually two possibilities -

1. 4 small books out of 8 possibilities and 1 large book out of 2 possibilities  -

This is in 8C4*2C1 =  \frac{8!}{(4!*4!)}*2 = \frac{(5*6*7*8)}{(2*3*4)}*2 = 140  ways

2. 5 small books out of 8 possibilities -

This in 8C5 =  \frac{8!}{(5!*3!)} = \frac{(6*7*8)}{(2*3)} = 56  ways.

So a total of (1) or (2)... using the OR-RULE

=140 + 56 = 196 ways.

Hence  Option (D) is our bet.

UNSTOPPABLE12 , Hope it is clear now.

Let me know if not.

gmexamtaker1 · Answer

Perfect explanation, indeed i didnt take into account the two different scenarios, thank you for your time  Gladiator59

Gladiator59 · Answer

Just to drill it further -

If you using the slotting technique ( as you pointed out in your PM) you could go ahead by this method -

1. First case (4 small and one large) -

Choosing small books first  - 8 ways to choose the first, 7 ways to choose the next, ... 5 ways to choose the 4th.

So (8*7*6*5) / 4! ( we divide by 4! as the order of books is not important)

and   choosing one large book in 2 ways.

Hence  70*2 = 140  ways.

2. Second case : Now slotting to choose 5 small books -

(8*7*6*5*4) ways to choose the five books and divide by 5! since the order is not important.

Hence in  \frac{(8*7*6*5*4)}{5!} = \frac{(8*7*6*5*4)}{(2*3*4*5)}= 56  ways

And then add the two as per OR -rule

Hence a total of 196. By either method.

ScottTargetTestPrep · Answer

Solution:
 
The number of ways he can take exactly 1 large book (along with 4 small books) is to choose 1 large book out of 2 available and 4 small books out of 8 available. Thus, we have: 2C1 x 8C4 = 2 x (8 x 7 x 6 x 5)/(4 x 3 x 2) = 2 x 7 x 2 x 5 = 140 ways.

The number of ways he can take no large book (and thus all 5 small books) is 2C0 x 8C5 = 1 x (8 x 7 x 6 x 5 x 4)/(5 x 4 x 3 x 2) = 8 x 7 = 56.

Therefore, the total number of ways he can take at most 1 large book with him on the trip is 140 + 56 = 196.

Answer: D

Kinshook · Answer

Given: Romi has a collection of 10 distinct books (8 small and 2 large). 
Asked: In how many ways can he select 5 books to take with him on a trip if he has room for only 1 large book?

Number of ways = 2C1*8C4 + 2C0*8C5 = 2* 70 + 56 = 196

IMO D

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