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25 Jun 2021, 08:15
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B
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Difficulty:
85%
(hard)
Question Stats:
48% (02:08) correct
52%
(02:33)
wrong
based on 230
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A shop sells 5 different types of sweets. In how many different ways a total of 8 sweets can be purchased?
(A) 125
(B) 495
(C) 795
(D) 840
(E) 930
(A) 125
(B) 495
(C) 795
(D) 840
(E) 930
01 Nov 2024, 02:39
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Alec0
A shop sells 5 different types of sweets. In how many different ways a total of 8 sweets can be purchased?
(A) 125
(B) 495
(C) 795
(D) 840
(E) 930
We have 5 different types of sweets: a, b, c, d, and e.
"Stars and Bars" method:
Imagine 8 stars in a row. To divide these stars among 5 types of sweets, we need 4 bars to create 5 separate sections. For example:
***|**|*|*|* represents purchasing 3 of sweet a, 2 of sweet b, 1 of sweet c, 1 of sweet d, and 1 of sweet e.
||****|**|** represents purchasing 0 of sweet a, 0 of sweet b, 4 of sweet c, 2 of sweet d, and 2 of sweet e.
|||*******| represents purchasing 0 of sweet a, 0 of sweet b, 0 of sweet c, 8 of sweet d, and 0 of sweet e.
||****|**|** represents purchasing 0 of sweet a, 0 of sweet b, 4 of sweet c, 2 of sweet d, and 2 of sweet e.
|||*******| represents purchasing 0 of sweet a, 0 of sweet b, 0 of sweet c, 8 of sweet d, and 0 of sweet e.
Thus, the problem becomes arranging these 8 identical stars and 4 identical bars in a row, which is calculated as 12!/(8! * 4!) = 495.
Answer: B.
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Hope this helps.
General Discussion
27 Jun 2021, 07:52
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Bunuel
Quickest way : Use partition method and the formula involved.
The equation can be written as \(a+b+c+d+e=8\), and we have find non negative solutions for the same
If there are 4 partitions, the answer would be (8+4)C4=\(\frac{12!}{8!/4!}=\frac{12*11*10*9}{4*3*2}=495\)
You can also calculate each type
B
