The toys are all identical. So, all you are trying to do is to choose the NUMBER of toys that get put in each box.
The condition given says that you must put at least one toy in each box; so, you have 8 identical toys remaining, and can distribute them between the three boxes in any way you like.
This can be thought of using a 'stars and bars' argument: any arrangement of these 8 toys is equivalent to a sequence of 8 $*$'s and 2 $|$'s, where each $*$ represents a toy and the $|$'s represent the cutoffs between boxes. For instance, $$ **|*****|* $$ represents the first box getting two of these eight toys, the second getting five, and the third getting one.
The number of such sequences is $\binom{10}{2}$: there are 10 positions total, and we need to choose which 2 of those positions are bars.
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Re: In how many ways 11 identical marbles be placed in 3 distinct jars suc [#permalink]
Bunuel wrote:
In how many ways 11 identical marbles be placed in 3 distinct jars such that no jar is empty?A. 72B. 54C. 45D. 42E. 36
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a+b+c = 11Total positive integer solutions = (n-1)C(r-1)Here, n = 11, r = 3So total Integer solutions = (11-1)C(3-1) = 10C2 = 45Answer: Option C _________________
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Re: In how many ways 11 identical marbles be placed in 3 distinct jars suc [#permalink]
Bunuel wrote:
In how many ways 11 identical marbles be placed in 3 distinct jars such that no jar is empty?A. 72B. 54C. 45D. 42E. 36
Are You Up For the Challenge: 700 Level Questions
I have been doing such problems by applying 2 equations.1) when we can distribute 0- n things to r boxes , no of ways = (n+r-1)C(r-1)2) when we are not allowed to keep any box empty =(n-1)C(r-1)Applying the second formula, 10C2 =45.C is the answer.It would be appreciated if someone can elaborate those two formulas.
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In how many ways 11 identical marbles be placed in 3 distinct jars suc [#permalink]
Bunuel wrote:
In how many ways 11 identical marbles be placed in 3 distinct jars such that no jar is empty?A. 72B. 54C. 45D. 42E. 36
Are You Up For the Challenge: 700 Level Questions
Explanation - How do we get these formulae?
Let the 11 identical marbles be denoted as M M M M M M M M M M MWe need to put 2 partitions so that we divide the marbles in 3 groups, say for example: M M M | M M M M | M M M M (where | represents the partition)Let us represent the above as: M M M P M M M M P M M M M
This is a 13 letter word with 11 Ms and 2 Ps and if we arrange it, we will get all distributions based on the position of the Ps
Ex:M M M P M M M M P M M M M => 3, 4, 4 distribution
M M M M M M M M M M P M P => 10, 1, 0 distribution; and so onNote: assigning 0 to a group is allowedThus, total arrangements = 13!/(11!2!) = 13C2 = (11+3-1)C(3-1)
Thus, non-negative integer solutions for a+b+c+d+... (r groups) = N is: (N+r-1)C(r-1) where N represents identical objects and r represents distinct groups
Let us now analyse "positive integer solutions":
Continuing in the same manner as above: Let the 11 identical marbles be denoted as M M M M M M M M M M MWe need to put 2 partitions so that we divide the marbles in 3 groupsTo ensure that no group ends up getting zero marbles, we assign 1 marble to each group first, leaving us with 8 marbles. We need to divide the marbles in 3 groups, say for example: M M M | M M M M | M (where | represents the partition)Let us represent the above as: M M M P M M M M P M
This is a 10 letter word with 8 Ms and 2 Ps and if we arrange it, we will get all distributions based on the position of the Ps
Thus, total arrangements = 10!/(8!2!) = 10C2 = (11-1)C(3-1)Thus, positive integer solutions for a+b+c+d+... (r groups) = N is: (N-1)C(r-1) where N represents identical objects and r represents distinct groups
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Originally posted by sujoykrdatta on 08 May 2020, 16:37.
Last edited by sujoykrdatta on 08 May 2020, 16:57, edited 1 time in total.
Re: In how many ways 11 identical marbles be placed in 3 distinct jars suc [#permalink]
minustark wrote:
Bunuel wrote:
In how many ways 11 identical marbles be placed in 3 distinct jars such that no jar is empty?A. 72B. 54C. 45D. 42E. 36
Are You Up For the Challenge: 700 Level Questions
I have been doing such problems by applying 2 equations.1) when we can distribute 0- n things to r boxes , no of ways = (n+r-1)C(r-1)2) when we are not allowed to keep any box empty =(n-1)C(r-1)Applying the second formula, 10C2 =45.C is the answer.It would be appreciated if someone can elaborate those two formulas.
I've always been using only (n+r-1)C(r-1) but when no box can be left empty, I put one marble in each box and apply the same formula for what is left.So in this case I put one marble in each jar so I'm left with 8 and so n=8, r=3, (n+r-1)C(r-1) = 10C2=45Posted from my mobile device
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Re: In how many ways 11 identical marbles be placed in 3 distinct jars suc [#permalink]
minustark wrote:
Bunuel wrote:
In how many ways 11 identical marbles be placed in 3 distinct jars such that no jar is empty?A. 72B. 54C. 45D. 42E. 36
Are You Up For the Challenge: 700 Level Questions
I have been doing such problems by applying 2 equations.1) when we can distribute 0- n things to r boxes , no of ways = (n+r-1)C(r-1)2) when we are not allowed to keep any box empty =(n-1)C(r-1)Applying the second formula, 10C2 =45.C is the answer.It would be appreciated if someone can elaborate those two formulas.
minustark
Please watch the following video in which I have explained where does this property come fromI hope it help!
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Re: In how many ways 11 identical marbles be placed in 3 distinct jars suc [#permalink]
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