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Combinations with at least 1 paperback = (all possible combinations) - (combinations with no paperbacks).NandishSS wrote:There are 8 books in a shelf that consist of 2 paperback books and 6 hardback books. If 4 books are selected at random without replacement, how many different combinations are there that at least one paperback book is selected?
A. 40
B. 45
C. 50
D. 55
E. 60
It must be true that:NandishSS wrote:There are 8 books in a shelf that consist of 2 paperback books and 6 hardback books. If 4 books are selected at random without replacement, how many different combinations are there that at least one paperback book is selected?
A. 40
B. 45
C. 50
D. 55
E. 60
OA:D
This looks like the slot approach but it has combinations in it. I though we have to divide by 2! to get rid of double countong. Obviously I'm wrong but I don't get when to divide at the end and when not toMatt@VeritasPrep wrote:We could also do
(Exactly One Paperback) + (Exactly Two Paperbacks) =>
(2 choose 1)*(6 choose 3) + (2 choose 2)*(6 choose 2) =>
2*20 + 1*15 =>
55
The issue is whether the items selected are interchangeable. Notice that in Matt's solution, he found 6 choose 3, by doing (6*5*4)/3!. The 3! accounts for the interchangeability of the items selected. (Choosing Lolita, 1984, and David Copperfield is the same as choosing 1984, David Copperfield and Lolita.) Similarly when he calculated 6 choose 2, he did (6*5)/2! Again, 2! accounts for the fact that the books are interchangeable. So 1) you can use the slot method, and 2) you are dividing the interchangeable components to avoid double-counting.hoppycat wrote:This looks like the slot approach but it has combinations in it. I though we have to divide by 2! to get rid of double countong. Obviously I'm wrong but I don't get when to divide at the end and when not toMatt@VeritasPrep wrote:We could also do
(Exactly One Paperback) + (Exactly Two Paperbacks) =>
(2 choose 1)*(6 choose 3) + (2 choose 2)*(6 choose 2) =>
2*20 + 1*15 =>
55
sweet thanks!DavidG@VeritasPrep wrote:The issue is whether the items selected are interchangeable. Notice that in Matt's solution, he found 6 choose 3, by doing (6*5*4)/3!. The 3! accounts for the interchangeability of the items selected. (Choosing Lolita, 1984, and David Copperfield is the same as choosing 1984, David Copperfield and Lolita.) Similarly when he calculated 6 choose 2, he did (6*5)/2! Again, 2! accounts for the fact that the books are interchangeable. So 1) you can use the slot method, and 2) you are dividing the interchangeable components to avoid double-counting.hoppycat wrote:This looks like the slot approach but it has combinations in it. I though we have to divide by 2! to get rid of double countong. Obviously I'm wrong but I don't get when to divide at the end and when not toMatt@VeritasPrep wrote:We could also do
(Exactly One Paperback) + (Exactly Two Paperbacks) =>
(2 choose 1)*(6 choose 3) + (2 choose 2)*(6 choose 2) =>
2*20 + 1*15 =>
55
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