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16 Dec 2009, 21:30
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Hello Everyone,
I'm confused about question #40 on the GClub Diag.
The problem and solution is as follows:
Romi has a collection of 10 distinct books (8 small and 2 large). 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?
(a) 56
(b) 126
(c) 152
(d) 196
(e) 252
Solution:
The number of ways 5 books can be chosen (with max. 1 large book) equals (total number of ways 5 books can be chosen) - (2 large + 3 small books will be chosen).
Total number of ways 5 books can be chosen \(= C_{10}^5 = \frac{10*9*8*7*6*5!}{5!5!} = 252\) .
Number of possible ways that 3 small and 2 large books will be chosen \(= 1*C_8^3 = 1 * 56 = 56\) .
Desired number of ways that 5 books can be taken \(= 252 - 56 = 196\) .
The correct answer is D.
Sorry, the solution is a little jumbled up from being pasted in. Anyway, can someone explain why the way I did the problem is wrong?
If only 1 large book can be chosen from 2 large books, then basically there are 2 possible ways for choosing the large book. (x2)
There are 8 small books that can be chosen for 4 small book "slots". That's like AAAABBBB, which is 8!/(4!*4!) which equals 70.
So then, my solution was 70 x 2 = 140. Please help! Ty!
I'm confused about question #40 on the GClub Diag.
The problem and solution is as follows:
Romi has a collection of 10 distinct books (8 small and 2 large). 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?
(a) 56
(b) 126
(c) 152
(d) 196
(e) 252
Solution:
The number of ways 5 books can be chosen (with max. 1 large book) equals (total number of ways 5 books can be chosen) - (2 large + 3 small books will be chosen).
Total number of ways 5 books can be chosen \(= C_{10}^5 = \frac{10*9*8*7*6*5!}{5!5!} = 252\) .
Number of possible ways that 3 small and 2 large books will be chosen \(= 1*C_8^3 = 1 * 56 = 56\) .
Desired number of ways that 5 books can be taken \(= 252 - 56 = 196\) .
The correct answer is D.
Sorry, the solution is a little jumbled up from being pasted in. Anyway, can someone explain why the way I did the problem is wrong?
If only 1 large book can be chosen from 2 large books, then basically there are 2 possible ways for choosing the large book. (x2)
There are 8 small books that can be chosen for 4 small book "slots". That's like AAAABBBB, which is 8!/(4!*4!) which equals 70.
So then, my solution was 70 x 2 = 140. Please help! Ty!
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16 Dec 2009, 22:33
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You're leaving out the possibility that he chooses not to bring a big book at all!
16 Dec 2009, 23:33
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Ah! Thanks you so much Eli. Yeah, that totally did not occur to me. Thanks!
