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09 Jan 2010, 10:08
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
83% (01:16) correct
18%
(01:37)
wrong
based on 40
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Responding to pm.
Customer can choose 3 out of 5 flavors 5C3=10;
Customer can choose 1 out of 2 cone type 2C1=2;
Total # of combinations possible 10*2=20.
Hope it's clear. The question would be harder if it were that customer can choose not only different flavors. And even more harder if customer can choose not only different flavors but less than three flavors too. You can try to solve these examples to master in combinatorics and permutation.
Quote:
Customer can choose 3 out of 5 flavors 5C3=10;
Customer can choose 1 out of 2 cone type 2C1=2;
Total # of combinations possible 10*2=20.
Hope it's clear. The question would be harder if it were that customer can choose not only different flavors. And even more harder if customer can choose not only different flavors but less than three flavors too. You can try to solve these examples to master in combinatorics and permutation.
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15 Aug 2013, 09:55
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At a particular ice-cream parlor, customers can choose among five different ice-cream flavors and may choose either a sugar cone or a waffle cone. Considering both ice-cream flavor and cone type, how many distinct triple-scoop cones with three different flavors are available?
(A) 20
(B) 10
(C) 15
(D) 18
(E) 25
(A) 20
(B) 10
(C) 15
(D) 18
(E) 25
Asifpirlo
We can select 3 out of 5 flavours in \(5C3\) ways
and can select cone in \(2C1\) ways
hence total ways = \(5C3*2C1 =20\)
hence A
