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A shop sells 5 different types of chocolates. In how many di 21 Dec 2019, 09:59
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95% (hard)

Question Stats:

41% (02:05) correct 59% (02:18) wrong based on 97 sessions

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A shop sells 5 different types of chocolates. In how many different ways a total of 8 chocolates can be purchased ?
A. 125
B. 495
C. 795
D. 840
E. 930

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B
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A shop sells 5 different types of chocolates. In how many di 22 Dec 2019, 03:11
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n+r-1 C r-1 ---- unequal distribution of 5 different chocolates to 8 ppl.

r= 5 n =8 -----> 12C4 = B
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A shop sells 5 different types of chocolates. In how many di 22 Dec 2019, 03:32
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a+b+c+d+e=8

a|b|c|d|e

We have to arrange 8 chocolates and 4 partitions

Total number of ways= \(\frac{(8+4)!}{8!*4!}\)= \(\frac{12*11*10*9}{4*3*2*1}\)= 495


Dillesh4096
A shop sells 5 different types of chocolates. In how many different ways a total of 8 chocolates can be purchased ?
A. 125
B. 495
C. 795
D. 840
E. 930

Posted from my mobile device
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A shop sells 5 different types of chocolates. In how many di 21 Dec 2019, 20:46
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Dillesh4096
A shop sells 5 different types of chocolates. In how many different ways a total of 8 chocolates can be purchased ?
A. 125
B. 495
C. 795
D. 840
E. 930

Posted from my mobile device
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A simple way would be to check different combinations

1) all of same type -
all 5 of any of 8 chocolates, so 5 ways

2) two types-
7,1 ; 6,2 ; 5,3 ; so \(5*4*3=60\) and
4,4 so \(\frac{5*4}{2}=10\)...
total 60+10=70

3) three types-
5,2,1 ; 4,3,1, so \(5*4*3*2=120\) and
6,1,1 ; 4,2,2 ; 3,3,2 so \(\frac{5*4*3}{2}*3=90\)...
total 210

4) 4 types -
4,2,1,1 ; 3,2,2,1, so \(\frac{5*4*3*2}{2}*2=120,\) and
3,3,1,1 so \(\frac{5*4*3*2}{(2*2)}=30\), and
5,1,1,1, so \(\frac{5*4*3*2}{3!}=20\), and
2,2,2,2, so \(\frac{5!}{4!}=5\)...
total 175

5) 5 types - 4,1,1,1,1, so \(\frac{5*4*3*2}{4!}=5\), and
3,2,1,1,1, so \(\frac{5*4*3*2}{3!}=20,\) and
2,2,2,1,1, so \(\frac{5!}{(3!2!})=10\),
so 35 ways

Total = 5+70+210+175+35=495

B
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A shop sells 5 different types of chocolates. In how many di 28 Jun 2024, 07:40
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