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In how many ways can 5 apples (identical) be distributed among 4 child
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29 Sep 2015, 08:18
Question Stats:
56% (01:39) correct 44% (01:59) wrong based on 190 sessions
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In how many ways can 5 apples (identical) be distributed among 4 child
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30 Sep 2015, 11:40
Bunuel wrote: In how many ways can 5 apples (identical) be distributed among 4 children? (Some children may get no apples.)
(A) 56 (B) 144 (C) 200 (D) 256 (E) 312 Number of ways of dividing 'n' identical objects into 'r' groups such that each group can contain any number of objects is given by \(n+r1_C_{r1}\) So, The number of ways of dividing 5 apples among 4 children is \(5+41_C_{41}\) = \(8_{C_3}\) = \(\frac{8!}{5!*3!}\) = \(56\) Answer: A




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Re: In how many ways can 5 apples (identical) be distributed among 4 child
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29 Sep 2015, 12:27
The answer is A. Here is how I found the answer Suppose we have five letter A’s representing apples. Let us also use three *’s which will represent partitions (four) between the apples belonging to different children. We order the A’s and *’s as we like and interpret all A’s before the first * as being apples belonging to Kathy. Now if you, the different ways these As and *s can be arranged, you will notice that each arrangement corresponds to a specific distribution. There are 5 As and 3 *s. The number of ways you can arrange them is [8!/(3!5!)] Generalised form for n = number of identical objects, and r = number of children is n+r1Cr1.



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Re: In how many ways can 5 apples (identical) be distributed among 4 child
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30 Sep 2015, 05:22
Let's just represent apple by the letter A and assume that in order to distribute these apples to different people, we just need to partition them in to 4 baskets.
In order to partition the apples in to 4 baskets, we would need three Partition bars , as shown below:
A  A  AA  A Child 1 Child 2 Child 3 Child 4
So basically we have 3 identical partition bars and 5 identical apples, how many ways can we arrange these?
8!/5!3! = 56



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Re: In how many ways can 5 apples (identical) be distributed among 4 child
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30 Sep 2015, 11:05
Hi camlan1990, This question comes with a bit of a 'twist.' Since the apples are IDENTICAL, you have to be careful about duplicate 'options'... For example: Giving 1 apple to child A and 1 apple to child B is the SAME as.... Giving 1 apple to child B and 1 apple to child A Since those duplicate options are NOT supposed to be counted twice, the Combination Formula is necessary (the solutions by icefrog and ankuragarwal1301 showcase that math). GMAT assassins aren't born, they're made, Rich
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Re: In how many ways can 5 apples (identical) be distributed among 4 child
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01 Oct 2015, 16:13
kunal555 wrote: Bunuel wrote: In how many ways can 5 apples (identical) be distributed among 4 children? (Some children may get no apples.)
(A) 56 (B) 144 (C) 200 (D) 256 (E) 312 Number of ways of dividing 'n' identical objects into 'r' groups such that each group can contain any number of objects is given by \(n+r1_C_{r1}\) So, The number of ways of dividing 5 apples among 4 children is \(5+41_C_{41}\) = \(8_{C_3}\) = \(\frac{8!}{5!*3!}\) = \(56\) Answer: ADoes this specifically include zero, as in zero apples in a basket? If so, what formula would you use in the event that everyone had to have at least 1 apple? Can you merely reduce "n" by 1, since n cant be zero? Suppose this there the case with 20 apples and 4 baskets.



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In how many ways can 5 apples (identical) be distributed among 4 child
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02 Oct 2015, 04:00
ar500 wrote: kunal555 wrote: Bunuel wrote: In how many ways can 5 apples (identical) be distributed among 4 children? (Some children may get no apples.)
(A) 56 (B) 144 (C) 200 (D) 256 (E) 312 Number of ways of dividing 'n' identical objects into 'r' groups such that each group can contain any number of objects is given by \(n+r1_C_{r1}\) So, The number of ways of dividing 5 apples among 4 children is \(5+41_C_{41}\) = \(8_{C_3}\) = \(\frac{8!}{5!*3!}\) = \(56\) Answer: ADoes this specifically include zero, as in zero apples in a basket? If so, what formula would you use in the event that everyone had to have at least 1 apple? Can you merely reduce "n" by 1, since n cant be zero? Suppose this there the case with 20 apples and 4 baskets. Yes, the above formula includes zero. The number of ways of distributing 'n' identical objects amongst 'r' groups such that each group gets at least 1 object is given by \({n1}_C_{r1}\) The number of ways of distributing 20 apples among 4 baskets such that each basket has at least 1 apple is \({201}_C_{41}\) = \({19}_C_{3}\)



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In how many ways can 5 apples (identical) be distributed among 4 child
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02 Oct 2015, 04:27
Quote: In how many ways can 5 apples (identical) be distributed among 4 children? (Some children may get no apples.)
(A) 56 (B) 144 (C) 200 (D) 256 (E) 312 I know a simple way to solve this problem. Consider it as an equation : A + B + C + D = 5. Where letters represent children and 5 = Apples. Just apply the formula : n+r1Cr1 where r = children and N = 5 i.e Apples. we will get answer as 56. Remember : This formula is used because question mentions that some children may get 0 apple. If this was not mentioned, there is a different formula to solve the question.



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Re: In how many ways can 5 apples (identical) be distributed among 4 child
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02 Oct 2015, 21:21
Bunuel wrote: In how many ways can 5 apples (identical) be distributed among 4 children? (Some children may get no apples.)
(A) 56 (B) 144 (C) 200 (D) 256 (E) 312 This questions can be solved by a property that's useful for calculating WHOLE NUMBER Solutions (All NonNegative Solution)
Whole Solution of a linear Equation a+b+c+d+e+.... = n
where we have r variables (a, b, c, d, e)
is (n+r1)C(r1) Here we have, a+b+c+d = 5 so Whole no. solution of the equation = (5+41)C(41) = 8C3 = 56 Answer: option A
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Re: In how many ways can 5 apples (identical) be distributed among 4 child
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22 Apr 2018, 08:54
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