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24 Jul 2017, 23:48
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
64% (02:42) correct
36%
(02:48)
wrong
based on 275
sessions
History
Date
Time
Result
Not Attempted Yet
Alfred, ever hungry, decides to order 4 desserts after his meal. If there are 7 types of pie and 8 types of ice cream from which to choose, and Alfred will have at most two types of ice cream, how many distinct groups of desserts could he consume in his post-prandial frenzy?
A. 588
B. 868
C. 903
D. 1806
E. 2010
A. 588
B. 868
C. 903
D. 1806
E. 2010
25 Jul 2017, 00:01
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Given data: There are 7 pies and 8 ice-creams to choose from.
Since Alfred decides to go with at most 2 ice-creams, out of the total 4 deserts.
He can have 0,1 or 2 ice-cream.
If he decides to have no ice cream :
Number of ways of choosing dessert(4 pies) : \(7c4 = \frac{7*6*5}{3*2}\) = 35 ways
If he decides to have 1 ice cream :
Number of ways of choosing dessert(3 pies and 1 ice-cream) : \(7c3*8c1 = \frac{7*6*5}{3*2} * 8 = 270\) ways
If he decides to have 2 ice cream :
Number of ways of choosing dessert(2 pies and 2 ice-cream) : \(7c2*8c2 = 49 * 12 = 588\) ways
Total ways in which Alfred choose dessert - \(35 + 270 + 588 = 903\)(Option C)
Since Alfred decides to go with at most 2 ice-creams, out of the total 4 deserts.
He can have 0,1 or 2 ice-cream.
If he decides to have no ice cream :
Number of ways of choosing dessert(4 pies) : \(7c4 = \frac{7*6*5}{3*2}\) = 35 ways
If he decides to have 1 ice cream :
Number of ways of choosing dessert(3 pies and 1 ice-cream) : \(7c3*8c1 = \frac{7*6*5}{3*2} * 8 = 270\) ways
If he decides to have 2 ice cream :
Number of ways of choosing dessert(2 pies and 2 ice-cream) : \(7c2*8c2 = 49 * 12 = 588\) ways
Total ways in which Alfred choose dessert - \(35 + 270 + 588 = 903\)(Option C)
Updated on: 25 Jul 2017, 04:05
Originally posted by TimeTraveller on 25 Jul 2017, 00:58.
Last edited by TimeTraveller on 25 Jul 2017, 04:05, edited 1 time in total.
Last edited by TimeTraveller on 25 Jul 2017, 04:05, edited 1 time in total.
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2 ice-creams + 2 pies = 8c2 * 7c2 = 28 * 21 = 588
1 ice-cream + 3 pies = 8c1 * 7c3 = 8 * 35 = 280
0 ice-cream + 4 pies = 7c4 = 35
No. of desserts = 903. Ans - C.
1 ice-cream + 3 pies = 8c1 * 7c3 = 8 * 35 = 280
0 ice-cream + 4 pies = 7c4 = 35
No. of desserts = 903. Ans - C.
