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11 Feb 2021, 06:15
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Difficulty:
65%
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Question Stats:
57% (02:01) correct
43%
(01:42)
wrong
based on 178
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An ice cream parlor has ice creams in eight different varieties. Number of ways of choosing 3 ice creams taking at least two ice creams of the same variety, is (Assume that ice creams of the same variety to be identical & available in unlimited supply)
(A) 56
(B) 64
(C) 100
(D) 105
(D) 110
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 56
(B) 64
(C) 100
(D) 105
(D) 110
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
11 Feb 2021, 22:18
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No of ways chosing the \(3\) ice cream of same varity = \(^8P_1 = 8\)
No of ways for chosing the \(2\) ice cream of one varity and the \(3^{rd}\) ice-cream of other varity = \(^8P_1\times ^7P_1=8\times 7=56\)
Therefore, no of ways \(= 8+56=64\)
Ans B
No of ways for chosing the \(2\) ice cream of one varity and the \(3^{rd}\) ice-cream of other varity = \(^8P_1\times ^7P_1=8\times 7=56\)
Therefore, no of ways \(= 8+56=64\)
Ans B
04 Jun 2025, 14:07
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Total ways to choose 3 ice creams (with repetitions allowed)
We use combinations with repetition to count all possible ways to choose 3 scoops from 8 varieties, assuming unlimited supply and identical scoops within each variety.
C = (8+3-1)!/3!(8-1)
C = 120
Subtract the number of combinations where all three ice creams are of different varieties. Just a simple and regular combination.
C = 8!/3!(5!)
C = 56
Total: 120 - 56 = 64.
Letter B.
We use combinations with repetition to count all possible ways to choose 3 scoops from 8 varieties, assuming unlimited supply and identical scoops within each variety.
C = (8+3-1)!/3!(8-1)
C = 120
Subtract the number of combinations where all three ice creams are of different varieties. Just a simple and regular combination.
C = 8!/3!(5!)
C = 56
Total: 120 - 56 = 64.
Letter B.
