D01-39

See example in the ebook from Magoosh team under Combinations: https://magoosh.resources.s3.amazonaws.c ... _eBook.pdf

C 5 over 8 = 8!/5!(8-5)! = 8*7*6*5*4*3*2*1/(5*4*3*2*1)*(3*2*1)
Simplify and you get 8*7*6/3*2*1
Simplify more and you get 4*7*2 = 8*7 = 56

You do the same with the other part of the equation.
So this for the first part. Hope this helps everyone!

Alternative solution
Since Romi can take at most 1 large book then he can take with him : All possible ways - (2 Large books and 3 books from 5 possible) =(5 small books from 10) - (3 small books from 8)=C510-C38=196

Hi there, i am relatively new to this group and could you please help me with the basics of this topic (may be with some link), i find this really difficult.

This is a great question. I did not think about the possibility of 5 small books. Bookmarked for future practice.

Alternative Solution:

I like to run mine backwards.

(Total Possibilities) \(-\) (Unwanted Outcomes)
Total Possibilities = 10C5 = 252
Unwanted Outcomes = Three Small Books, both Large Books
Three small books: 8C3 = 56
Both large books: 2C2 = 1

= 10C5 \(-\) 8C3(2C2)
= 252 \(-\) 56(1)
= 252 \(-\) 56
= 196 (Wanted Outcomes)

Hello, taha1234. Since order does not matter, we should not use a permutation. Consider a valid combination of small {a, b, c, d} and large books {E} with a change of order tossed in:

abcdE
abcEd
abEcd
aEbcd
Eabcd

Did the order in which book E (or any other book) was selected affect which books on the whole were selected? Not at all. I could rewrite different combinations of the same five letters a hundred times over (120 times, to be exact), but the books selected would be the same five books. That is how we know we are dealing with a combination.

I hope that helps clarify the issue for you or anyone else who stumbles across this thread with a similar question. Good luck with your studies.

- Andrew

Hey quick question,
This problem states that all the books are distinct.

so won there be much more cases? ( a permutation?)

Thanks

Hello, NikhilJose. I have touched on this very point two posts above yours, at the top of the page. In short, yes, we do have distinct books or titles, but notice that these books are separated into one of two categories, small and large. Within the combination formula, we can select, say, four small books and one large one, so the distinct titles do matter, but the order in which those four small books are selected does not matter, as I have illustrated in the earlier post. Hence, we know we are not dealing with a permutation.

I hope it makes sense. Good luck with your studies.

- Andrew

I think this is a high-quality question and I agree with explanation.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

I think this is a high-quality question and the explanation isn't clear enough, please elaborate.

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

Romi can either select 5 small books or 4 small books and 1 large book. Calculating the combinations, we have \(C^5_8\) for selecting 5 small books from 8, and \(C^4_8 * C^1_2\) for selecting 4 small and 1 large book. Adding these gives \(C^5_8 + C^4_8 * C^1_2 = 56 + 140 = 196\).

Answer: D.

Hope it's clear.