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In how many ways, 10 identical chocolates be distributed among 3 child
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Updated on: 07 Oct 2019, 00:27
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In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate? A. 36 B. 66 C. 72 D. 78 E. 84.
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Originally posted by Kjol on 06 Oct 2019, 22:06.
Last edited by Kjol on 07 Oct 2019, 00:27, edited 1 time in total.




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Re: In how many ways, 10 identical chocolates be distributed among 3 child
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07 Oct 2019, 09:44
Kjol wrote: In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?
A. 36 B. 66 C. 72 D. 78 E. 84. Another approach is to look for a patternSay the 3 children are A, B, C We'll denote each outcome as follows: # chocolates for A  # chocolates for B  # chocolates for C Number of outcomes in which child A receives exactly 1 chocolate1  1  8 1  2  7 1  3  6 1  4  5 1  5  4 1  6  3 1  7  2 1  8  1 8 outcomes Number of outcomes in which child A receives exactly 2 chocolates2  1  7 2  2  6 2  3  5 2  4  4 2  5  3 2  6  2 2  7  1 7 outcomes Number of outcomes in which child A receives exactly 3 chocolates3  1  6 3  2  5 3  3  4 3  4  3 3  5  2 3  6  1 6 outcomes See the pattern? So, the TOTAL number of outcomes = 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 Answer: A Cheers, Brent
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In how many ways, 10 identical chocolates be distributed among 3 child
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07 Oct 2019, 00:12
Kjol wrote: In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?
A. 36 B. 66 C. 72 D. 78 E. 84. Two ways.. Formula:First give 1 each to the three children, so we are left with 103=7. Now you can distribute these 7 to any child in \((n1)C7=9C7=\frac{9!}{2!7!}=36\) first give one to each, hence 3 gone. Now you can distribute the remaining 7 to anyone as the restriction of at least one is done.. A+B+C=7.... Now, take this as 7 marbles and 2 partitions, so total 9 things. You can choose these 2 partitions in 9C2 ways=\(\frac{9!}{7!2!}=\frac{9*8}{2}=36\) So, the formula \((n+k1)C(k1)\) comes from k items, so k1 partition and then choosing these k1 partition from it..Here n=7, and k=3..thus \((n+k1)C(k1)=(7+31)C(31)=9C2\) Calculations:Let the number each can get be 8,1,13!/2=3 ways 7,2,13! ways 6,3,13!6,2,2  3!/2=3 5,4,1  3! 5,3,2  3!4,4,2 3!/2=3 4,3,3  3!/2=3 Total = 4*3!+4*3=4*6+12=36 A
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Re: In how many ways, 10 identical chocolates be distributed among 3 child
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07 Oct 2019, 00:14
The possible combinations are: 118 > with 3 rearrangements ( because 1 is repeated, the possibilities = \(\frac{3!}{2!} = 3\)) 127 > with 6 rearrangements (because all numbers are different, the possibilities = \(3! = 6\)) 136 > with 6 rearrangements 145 > with 6 rearrangements 226 > with 3 rearrangements 235 > with 6 rearrangements 244 > with 3 rearrangements 334 > with 3 rearrangements Total = 36 Another faster way is the "Stars and Bars Theorem", which says that for n items distributed on k elements, where each of the elements must have a positive value, then the number of possibilities = (n1)C(k1). and in our case, it will be (101)C(31) = 9C2 = 36
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Re: In how many ways, 10 identical chocolates be distributed among 3 child
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07 Oct 2019, 09:10
Kjol wrote: In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?
A. 36 B. 66 C. 72 D. 78 E. 84. After distributing 1 chocolate to each child Remaining 7 chocolates are to be distributed among 3 children *** This can be done in (7+2)!/7!2! = 9!/7!2! = 9*8/2 = 36 IMO A
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Re: In how many ways, 10 identical chocolates be distributed among 3 child
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07 Oct 2019, 09:39
Kjol wrote: In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?
A. 36 B. 66 C. 72 D. 78 E. 84. Formula: Number of positive integral solutions of the equation \(x_1\) + \(x_2\) + . . . . + \(x_r\) = n is \((n1)c_{r1}\) We have to find \(x_1\) + \(x_2\) + \(x_3\) = 10 —> Number of positive integral solutions = \((101)c_{31}\) = \(9c_2\) = 9*8/2! = 36 IMO Option A Posted from my mobile device



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Re: In how many ways, 10 identical chocolates be distributed among 3 child
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07 Oct 2019, 19:24
chetan2u wrote: Kjol wrote: In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?
A. 36 B. 66 C. 72 D. 78 E. 84. Two ways.. Formula:First give 1 each to the three children, so we are left with 103=7. Now you can distribute these 7 to any child in \((n1)C7=9C7=\frac{9!}{2!7!}=36\) Calculations:Let the number each can get be 8,1,13!/2=3 ways 7,2,13! ways 6,3,13!6,2,2  3!/2=3 5,4,1  3! 5,3,2  3!4,4,2 3!/2=3 4,3,3  3!/2=3 Total = 4*3!+4*3=4*6+12=36 A Hello Chetan2u, Whats this formula approach? Can you please explain ? I was able to understand the calculation approach.



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In how many ways, 10 identical chocolates be distributed among 3 child
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11 Oct 2019, 06:25
Another Way to solve it is to make a list where you start each point of the list with how many chocolates you give to the first child, so the list is 8 7 6 usw.
Then you check in how many ways you can distribute the remaining chocolates among the other two children. For 8 it is 1 For 7 It is 2 usw, in the end you add all the numbers between 1 and 8 inclusive, and the sum is 36



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Re: In how many ways, 10 identical chocolates be distributed among 3 child
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13 Oct 2019, 01:31
chetan2u wrote: Kjol wrote: In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?
A. 36 B. 66 C. 72 D. 78 E. 84. Two ways.. Formula:First give 1 each to the three children, so we are left with 103=7. Now you can distribute these 7 to any child in \((n1)C7=9C7=\frac{9!}{2!7!}=36\) Calculations:Let the number each can get be 8,1,13!/2=3 ways 7,2,13! ways 6,3,13!6,2,2  3!/2=3 5,4,1  3! 5,3,2  3!4,4,2 3!/2=3 4,3,3  3!/2=3 Total = 4*3!+4*3=4*6+12=36 A Could you please explain why you're doing 9C7 instead of 10C7 after distributing one to each child?



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Re: In how many ways, 10 identical chocolates be distributed among 3 child
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17 Oct 2019, 14:05
chetan2u Can you please explain the formula approach



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Re: In how many ways, 10 identical chocolates be distributed among 3 child
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18 Oct 2019, 09:51
chetan2u wrote: Kjol wrote: In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?
A. 36 B. 66 C. 72 D. 78 E. 84. Two ways.. Formula:First give 1 each to the three children, so we are left with 103=7. Now you can distribute these 7 to any child in \((n1)C7=9C7=\frac{9!}{2!7!}=36\) Calculations:Let the number each can get be 8,1,13!/2=3 ways 7,2,13! ways 6,3,13!6,2,2  3!/2=3 5,4,1  3! 5,3,2  3!4,4,2 3!/2=3 4,3,3  3!/2=3 Total = 4*3!+4*3=4*6+12=36 A chetan2uwhy do we have to do "(n−1)" ?



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Re: In how many ways, 10 identical chocolates be distributed among 3 child
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26 Oct 2019, 21:34
Kjol wrote: In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?
A. 36 B. 66 C. 72 D. 78 E. 84. first give one to each, hence 3 gone. Now you can distribute the remaining 7 to anyone as the restriction of at least one is done.. A+B+C=7.... Now, take this as 7 marbles and 2 partitions, so total 9 things. You can choose these 2 partitions in 9C2 ways=\(\frac{9!}{7!2!}=\frac{9*8}{2}=36\) So, the formula (n+k1)C(k1) comes from k items, so k1 partition and then choosing these k1 partition from it.. Here n=7, and k=3..thus (n+k1)C(k1)=(7+31)C(31)=9C2 prgmatbiz ManjariMishra prabsahi and ShreyasJavaharHAPPY DIWALI !!!!
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Re: In how many ways, 10 identical chocolates be distributed among 3 child
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