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In how many ways, 10 identical chocolates be distributed among 3 child [#permalink]
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Question Stats: Hide Show timer StatisticsIn 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.
Originally posted by Kjol on 06 Oct 2019, 21:06.
Re: In how many ways, 10 identical chocolates be distributed among 3 child [#permalink]
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 pattern Say the 3 children are A, B, CWe'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 chocolate 1 | 1 | 81 | 2 | 71 | 3 | 61 | 4 | 51 | 5 | 41 | 6 | 31 | 7 | 21 | 8 | 18 outcomes Number of outcomes in which child A receives exactly 2 chocolates 2 | 1 | 72 | 2 | 62 | 3 | 52 | 4 | 42 | 5 | 32 | 6 | 22 | 7 | 17 outcomes Number of outcomes in which child A receives exactly 3 chocolates 3 | 1 | 63 | 2 | 53 | 3 | 43 | 4 | 33 | 5 | 23 | 6 | 16 outcomes See the pattern?So, the TOTAL number of outcomes = 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 Answer: ACheers, Brent _________________
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Math Expert Joined: 02 Aug 2009 Posts: 10653
In how many ways, 10 identical chocolates be distributed among 3 child [#permalink]
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 10-3=7.Now you can distribute these 7 to any child in \((n-1)C7=9C7=\frac{9!}{2!7!}=36\)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 \((n+k-1)C(k-1)\) comes from k items, so k-1 partition and then choosing these k-1 partition from it.. Here n=7, and k=3..thus \((n+k-1)C(k-1)=(7+3-1)C(3-1)=9C2\)
7,2,1--3! ways 5,4,1 -- 3!
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Re: In how many ways, 10 identical chocolates be distributed among 3 child [#permalink] 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 rearrangements145 --> with 6 rearrangements226 --> with 3 rearrangements235 --> with 6 rearrangements244 --> with 3 rearrangements334 --> with 3 rearrangementsTotal = 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 = (n-1)C(k-1). and in our case, it will be (10-1)C(3-1) = 9C2 = 36
Re: In how many ways, 10 identical chocolates be distributed among 3 child [#permalink]
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 childRemaining 7 chocolates are to be distributed among 3 children*|*|* This can be done in (7+2)!/7!2! = 9!/7!2! = 9*8/2 = 36IMO A _________________
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Re: In how many ways, 10 identical chocolates be distributed among 3 child [#permalink]
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 \((n-1)c_{r-1}\)We have to find \(x_1\) + \(x_2\) + \(x_3\) = 10—> Number of positive integral solutions = \((10-1)c_{3-1}\) = \(9c_2\) = 9*8/2! = 36IMO Option APosted from my mobile device
Senior Manager Joined: 09 Jun 2014 Posts: 260 Location: India Concentration: General Management, Operations
Re: In how many ways, 10 identical chocolates be distributed among 3 child [#permalink]
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 10-3=7.Now you can distribute these 7 to any child in \((n-1)C7=9C7=\frac{9!}{2!7!}=36\)
7,2,1--3! ways 5,4,1 -- 3! A Hello Chetan2u,Whats this formula approach? Can you please explain ?I was able to understand the calculation approach.
Intern Joined: 31 Oct 2018 Posts: 2 Location: Switzerland Schools: Tsinghua GPA: 3.25 WE:Sales (Consumer Products)
In how many ways, 10 identical chocolates be distributed among 3 child [#permalink] 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 is876 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
Manager Joined: 30 Sep 2019 Posts: 97 Location: India GMAT 1: 700 Q44 V40 GMAT 2: 700 Q49 V37 GMAT 3: 720 Q49 V38
Re: In how many ways, 10 identical chocolates be distributed among 3 child [#permalink]
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 10-3=7.Now you can distribute these 7 to any child in \((n-1)C7=9C7=\frac{9!}{2!7!}=36\)
7,2,1--3! ways 5,4,1 -- 3! 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 [#permalink] chetan2u Can you please explain the formula approach
Re: In how many ways, 10 identical chocolates be distributed among 3 child [#permalink]
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 10-3=7.Now you can distribute these 7 to any child in \((n-1)C7=9C7=\frac{9!}{2!7!}=36\)
7,2,1--3! ways 5,4,1 -- 3! A chetan2u why do we have to do "(n−1)" ?
Math Expert Joined: 02 Aug 2009 Posts: 10653
Re: In how many ways, 10 identical chocolates be distributed among 3 child [#permalink]
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+k-1)C(k-1) comes from k items, so k-1 partition and then choosing these k-1 partition from it..Here n=7, and k=3..thus (n+k-1)C(k-1)=(7+3-1)C(3-1)=9C2prgmatbiz ManjariMishra prabsahi and ShreyasJavahar HAPPY DIWALI !!!! _________________
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Re: In how many ways, 10 identical chocolates be distributed among 3 child [#permalink]
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. 10 identical chocolates and 3 different children getting at least 1:\(r=n:(x_1+1)+(x_2+1)+(x_3+1)=10\)\(r=n:(x_1)+(x_2)+(x_3)=10-3=7\)\(C(n+r-1;n,r-1)=\frac{9!}{7!2!}=36\)Ans (A)
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Re: In how many ways, 10 identical chocolates be distributed among 3 child [#permalink] Hello from the GMAT Club BumpBot!Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: In how many ways, 10 identical chocolates be distributed among 3 child [#permalink] 20 Dec 2021, 04:09 |
Taruna likes 10 different kinds of chocolates. in how many ways can she have at least one of them?
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