In how many ways can 16 different gifts be divided among four children if there is
no restriction on the number of gifts that each child can receive? In other words, a
child can receive as few as zero gifts and as many as 16 gifts?
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Rahul@gurome
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This type of problems can be solved by the famous "stick-separator" method. Which I have discussed in detail in another post, https://www.beatthegmat.com/oranges-t70953.html#320828.yellowho wrote:In how many ways can 16 different gifts be divided among four children if there is
no restriction on the number of gifts that each child can receive? In other words, a
child can receive as few as zero gifts and as many as 16 gifts?
The problem in the above mentioned post is slightly different, but the concept I used is applicable here too. Go through it and let me know if any further assistance is needed.
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yellowho
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Actually I applied your method originally and still got it wrong. I must be doing something wrong. Can you please take a look:
1) Since you can have zero gifts. there are 16 bars and 3 seperators and 4 regions that will make up the 4 children
2) 19!/3!16!
Answer is OA is 4^16 which also makes sense since there are 4 places each of the 16 items can be placed. My real question is what am I doing wrong the seperator method??
[quote="Rahul@gurome"][quote="yellowho"]In how many ways can 16 different gifts be divided among four children if there is
no restriction on the number of gifts that each child can receive? In other words, a
child can receive as few as zero gifts and as many as 16 gifts?[/quote]
This type of problems can be solved by the famous "stick-separator" method. Which I have discussed in detail in another post, [url]https://www.beatthegmat.com/oranges-t70953.html#320828[/url].
The problem in the above mentioned post is slightly different, but the concept I used is applicable here too. Go through it and let me know if any further assistance is needed.[/quote]
1) Since you can have zero gifts. there are 16 bars and 3 seperators and 4 regions that will make up the 4 children
2) 19!/3!16!
Answer is OA is 4^16 which also makes sense since there are 4 places each of the 16 items can be placed. My real question is what am I doing wrong the seperator method??
[quote="Rahul@gurome"][quote="yellowho"]In how many ways can 16 different gifts be divided among four children if there is
no restriction on the number of gifts that each child can receive? In other words, a
child can receive as few as zero gifts and as many as 16 gifts?[/quote]
This type of problems can be solved by the famous "stick-separator" method. Which I have discussed in detail in another post, [url]https://www.beatthegmat.com/oranges-t70953.html#320828[/url].
The problem in the above mentioned post is slightly different, but the concept I used is applicable here too. Go through it and let me know if any further assistance is needed.[/quote]
yellowho ,
In the question Rahul is talking about, ALL THE 16 GIFTS ARE CONSIDERED SAME.
where as your question deals with 16 DIFFERENT. GIFTS,
So just to clear your doubt:
16 different gifts => Ans 4^16,
16 identical gifts => ans 19!/3!16! .
In the question Rahul is talking about, ALL THE 16 GIFTS ARE CONSIDERED SAME.
where as your question deals with 16 DIFFERENT. GIFTS,
So just to clear your doubt:
16 different gifts => Ans 4^16,
16 identical gifts => ans 19!/3!16! .
Beautiful method - Rahul. Thanks. This is great.
In this specific case since the items are all "different"
I think it can be thought of as assigning the children to each distinct gift and not the other way around.
There are 16 different gifts sitting in a row. How many ways can the 4 children be "packed" into each of the gifts if there is no restriction on the number of gifts the children can get?
First gift: 4
Second gift: 4
Third gift: 4
.....
Sixteenth gift: 4
4 * 4 * 4 *.... => 4^16
Quick question for Rahul: In your stick method ---
If I put 16 different sticks (each standing for a gift) + 3 separators
Then why doesn't (19!)/3!*1! yield the same result... If the gifts are all different then its like dividing by 1! 16 times in the denominator, right? Just as you divide by 16! if the items are all identical. Why doesn't that work here?
Thanks again.
In this specific case since the items are all "different"
I think it can be thought of as assigning the children to each distinct gift and not the other way around.
There are 16 different gifts sitting in a row. How many ways can the 4 children be "packed" into each of the gifts if there is no restriction on the number of gifts the children can get?
First gift: 4
Second gift: 4
Third gift: 4
.....
Sixteenth gift: 4
4 * 4 * 4 *.... => 4^16
Quick question for Rahul: In your stick method ---
If I put 16 different sticks (each standing for a gift) + 3 separators
Then why doesn't (19!)/3!*1! yield the same result... If the gifts are all different then its like dividing by 1! 16 times in the denominator, right? Just as you divide by 16! if the items are all identical. Why doesn't that work here?
Thanks again.
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Rahul@gurome
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Thanks jaxis for pointing out the difference.
Here the gifts are different hence there is exist a much more intuitive and easy method as gmat1011 explained.
Say the 4 gifts are A, B, and C and the children are x and y.
Now according to simple application of "stick-separator" method, there are four possible combinations: (0, 3), (1, 2), (2, 1), and (3, 0).
But as the gifts are different, (1, 2) can have three possible sub-arrangements : (A, B + C), (B, A + C), and (C, A + B). Same goes for (2, 1). Hence a total of 8 different combinations.
Whereas if you try to modify the "stick-separator" method as you did, then the number is much more because it takes some false sub-arrangements into account. Such as it interprets (0, 3) as (0, A + B + C), (0, A + C + B), (0, B + A + C), (0, B + C + A), (0, C + A + B), and (0, C + B + A).
Hope this helps to clear the confusion.
Here the gifts are different hence there is exist a much more intuitive and easy method as gmat1011 explained.
Let's analyze for a small number of gifts, say 3 and 2 children.gmat1011 wrote:Quick question for Rahul: In your stick method ---
If I put 16 different sticks (each standing for a gift) + 3 separators
Then why doesn't (19!)/3!*1! yield the same result... If the gifts are all different then its like dividing by 1! 16 times in the denominator, right? Just as you divide by 16! if the items are all identical. Why doesn't that work here?
Say the 4 gifts are A, B, and C and the children are x and y.
Now according to simple application of "stick-separator" method, there are four possible combinations: (0, 3), (1, 2), (2, 1), and (3, 0).
But as the gifts are different, (1, 2) can have three possible sub-arrangements : (A, B + C), (B, A + C), and (C, A + B). Same goes for (2, 1). Hence a total of 8 different combinations.
Whereas if you try to modify the "stick-separator" method as you did, then the number is much more because it takes some false sub-arrangements into account. Such as it interprets (0, 3) as (0, A + B + C), (0, A + C + B), (0, B + A + C), (0, B + C + A), (0, C + A + B), and (0, C + B + A).
Hope this helps to clear the confusion.
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DarkKnight
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Hi Rahul,Rahul@gurome wrote:Thanks jaxis for pointing out the difference.
Here the gifts are different hence there is exist a much more intuitive and easy method as gmat1011 explained.
Let's analyze for a small number of gifts, say 3 and 2 children.gmat1011 wrote:Quick question for Rahul: In your stick method ---
If I put 16 different sticks (each standing for a gift) + 3 separators
Then why doesn't (19!)/3!*1! yield the same result... If the gifts are all different then its like dividing by 1! 16 times in the denominator, right? Just as you divide by 16! if the items are all identical. Why doesn't that work here?
Say the 4 gifts are A, B, and C and the children are x and y.
Now according to simple application of "stick-separator" method, there are four possible combinations: (0, 3), (1, 2), (2, 1), and (3, 0).
But as the gifts are different, (1, 2) can have three possible sub-arrangements : (A, B + C), (B, A + C), and (C, A + B). Same goes for (2, 1). Hence a total of 8 different combinations.
Whereas if you try to modify the "stick-separator" method as you did, then the number is much more because it takes some false sub-arrangements into account. Such as it interprets (0, 3) as (0, A + B + C), (0, A + C + B), (0, B + A + C), (0, B + C + A), (0, C + A + B), and (0, C + B + A).
Hope this helps to clear the confusion.
I can't visualize how we got to 4^16???? I don't understand what to make out of "I think it can be thought of as assigning the children to each distinct gift and not the other way around. " statement. Can you please explain how to get to 4^16? Thanks for your help.
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We have 16 different gifts and we have to distribute them among 4 children.DarkKnight wrote:I can't visualize how we got to 4^16???? I don't understand what to make out of "I think it can be thought of as assigning the children to each distinct gift and not the other way around. " statement. Can you please explain how to get to 4^16? Thanks for your help.
Now, each of the 16 gifts can be given to any one of the 4 children.
Hence each gift can be distributed in 4 ways.
Thus, 16 gifts can be distributed by 4^16 ways.
Hope it helps.
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Understood. I was focusing on groupings but not on counting method. Thanks for clearing it up.Rahul@gurome wrote:We have 16 different gifts and we have to distribute them among 4 children.DarkKnight wrote:I can't visualize how we got to 4^16???? I don't understand what to make out of "I think it can be thought of as assigning the children to each distinct gift and not the other way around. " statement. Can you please explain how to get to 4^16? Thanks for your help.
Now, each of the 16 gifts can be given to any one of the 4 children.
Hence each gift can be distributed in 4 ways.
Thus, 16 gifts can be distributed by 4^16 ways.
Hope it helps.
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Since there is no limitation on number of Gift for each kid. This makes it confusing.
1 kid can have 13 gifts & other 3 hold 1 gift each.
OR
1 kid can have 12 gifts & second kid has 2 gift & other 2 kids hold 1 gift each.
We need Mitch's help to solve this question.
1 kid can have 13 gifts & other 3 hold 1 gift each.
OR
1 kid can have 12 gifts & second kid has 2 gift & other 2 kids hold 1 gift each.
We need Mitch's help to solve this question.
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GMATGuruNY
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Let the 4 children be A, B, C and D.yellowho wrote:In how many ways can 16 different gifts be divided among four children if there is no restriction on the number of gifts that each child can receive? In other words, a child can receive as few as zero gifts and as many as 16 gifts?
Each gift must be assigned to a child.
Number of options for the 1st gift = 4. (A, B, C or D)
Number of options for the 2nd gift = 4. (A, B, C or D)
Number of options for the 3rd gift = 4. (A, B, C or D)
Until...
Number of options for the 16th gift = 4. (A, B, C or D)
To combine these options, we multiply:
4*4*4*...*4*4*4 = [spoiler]4¹�[/spoiler].
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gmatdriller
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What is the difference if the question had asked:
In how many ways can 4 children be assigned 16 different gifts...?
Assume a child can receive as few as 0 gifts and as many as 16.
Thanks
In how many ways can 4 children be assigned 16 different gifts...?
Assume a child can receive as few as 0 gifts and as many as 16.
Thanks
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Same problem.gmatdriller wrote:What is the difference if the question had asked:
In how many ways can 4 children be assigned 16 different gifts...?
Assume a child can receive as few as 0 gifts and as many as 16.
Thanks
Since it is possible for a child not to receive a gift, but each gift MUST be given to a child, we count from the perspective of the gifts:
Number of options for the 1st gift = 4. (Any of the 4 children.)
Number of options for the 2nd gift = 4. (Any of the 4 children.)
Number of options for the 3rd gift = 4. (Any of the 4 children.)
Until...
Number of options for the 16th gift = 4. (Any of the 4 children.)
To combine these options, we multiply:
4*4*4*...*4*4*4 = 4¹�.
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Take the task of distributing the 16 gifts and break it into stages.yellowho wrote:In how many ways can 16 different gifts be divided among four children if there is no restriction on the number of gifts that each child can receive? In other words, a child can receive as few as zero gifts and as many as 16 gifts?
Stage 1: Select which child gets the 1st gift
There are 4 children who can receive this gift, so we can complete stage 1 in 4 ways
Stage 2: Select which child gets the 2nd gift
There are 4 children who can receive this gift, so we can complete stage 2 in 4 ways
Stage 3: Select which child gets the 3rd gift
There are 4 children who can receive this gift, so we can complete stage 3 in 4 ways
.
.
.
Stage 16: Select which child gets the 16th gift
There are 4 children who can receive this gift, so we can complete stage 16 in 4 ways
By the Fundamental Counting Principle (FCP), we can complete all 16 stages (and thus distribute all 16 gifts) in (4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4) ways ([spoiler]= 4¹� ways[/spoiler])
--------------------------
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gmatdriller
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Thanks GmatGuru for the response.GMATGuruNY wrote:Same problem.gmatdriller wrote:What is the difference if the question had asked:
In how many ways can 4 children be assigned 16 different gifts...?
Assume a child can receive as few as 0 gifts and as many as 16.
Thanks
Since it is possible for a child not to receive a gift, but each gift MUST be given to a child, we count from the perspective of the gifts:
Number of options for the 1st gift = 4. (Any of the 4 children.)
Number of options for the 2nd gift = 4. (Any of the 4 children.)
Number of options for the 3rd gift = 4. (Any of the 4 children.)
Until...
Number of options for the 16th gift = 4. (Any of the 4 children.)
To combine these options, we multiply:
4*4*4*...*4*4*4 = 4¹�.
Because some children may not receive a gift, we cannot tell who it is.
Better to start from the perspective of the gifts.
But, In the event each child MUST receive at least a gift:
Would this be a reasonable question on the gmat?
Regards.
