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by yellowho » Tue Mar 01, 2011 9:54 pm
In how many ways can 6 identical coins be distributed among Alex, Bea and Chad? Note: Some people may receive zero coins.


I don't have solution to this. Is this 28?
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by Night reader » Tue Mar 01, 2011 10:29 pm
IOM [spoiler]because some of three people (ABC) may receive zero coins, which are identical
1) A(6) B(0) C(0)
2) A(5) B(1) C(0)
3) A(5) B(0) C(1)
4) A(4) B(2) C(0)
5) A(4) B(0) C(2)
6) A(4) B(1) C(1)
7) A(3) B(3) C(0)
8) A(3) B(0) C(3)
9) A(3) B(2) C(1)
10) A(3) B(1) C(2)
11) A(2) B(4) C(0)
12) A(2) B(0) C(4)
13) A(2) B(3) C(1)
14) A(2) B(2) C(2)
15) A(2) B(1) C(3)
16) A(1) B(5) (0)
17) A(1) B(0) C(5)
18) A(1) B(1) C(4)
19) A(1) B(4) C(1)
20) A(1) B(3) C(2)
21) A(1) B(2) C(3)
22) A(0) B(6) C(0)
23) A(0) B(0) C(6)
24) A(0) B(5) C(1)
25) A(0) B(1) C(5)
26) A(0) B(4) C(2)
27) A(0) B(2) C(4)
28) A(0) B(3) C(3)

added some more and we are equal - total 28 ways



[/spoiler]
yellowho wrote:In how many ways can 6 identical coins be distributed among Alex, Bea and Chad? Note: Some people may receive zero coins.


I don't have solution to this. Is this 28?
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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by BarryLi » Tue Mar 01, 2011 11:03 pm
[spoiler]Thank goodness the coins are identical.

For 0 coins, there is 1 combination: 0-0-0
For 1 coin, there are 3 combinations: 0-0-1, 0-1-0, 1-0-0
For 2 coins, there are 6 combinations: 0-0-2, 0-1-1, 0-2-0, 1-0-1, 1-1-0, 2-0-0

There may be a pattern of adding 1, 2, 3 to the total number of combinations, but I'll working out the case for 3 coins just to confirm this.

For 3 coins, there are 10 distributions: 0-0-3, 0-1-2, 0-2-1, 0-3-0, 1-0-2, 1-1-1, 1-2-0, 2-0-1, 2-1-0, 3-0-0

From the sequence 1, 3, 6, 10, the number of terms seems to be increasing by 2, 3, 4, etc.

0 coins: 1
1 coin: 3
2 coins: 6
3 coins: 10
4 coins: 15
5 coins: 21
6 coins: 28

Yes, the solution is 28.[/spoiler]
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by rohu27 » Tue Mar 01, 2011 11:21 pm
can we have a case here where a=b=c=0?
Night reader wrote:IOM [spoiler]because some of three people (ABC) may receive zero coins, which are identical
1) A(6) B(0) C(0)
2) A(5) B(1) C(0)
3) A(5) B(0) C(1)
4) A(4) B(2) C(0)
5) A(4) B(0) C(2)
6) A(4) B(1) C(1)
7) A(3) B(3) C(0)
8) A(3) B(0) C(3)
9) A(3) B(2) C(1)
10) A(3) B(1) C(2)
11) A(2) B(4) C(0)
12) A(2) B(0) C(4)
13) A(2) B(3) C(1)
14) A(2) B(2) C(2)
15) A(2) B(1) C(3)
16) A(1) B(5) (0)
17) A(1) B(0) C(5)
18) A(1) B(1) C(4)
19) A(1) B(4) C(1)
20) A(1) B(3) C(2)
21) A(1) B(2) C(3)
22) A(0) B(6) C(0)
23) A(0) B(0) C(6)
24) A(0) B(5) C(1)
25) A(0) B(1) C(5)
26) A(0) B(4) C(2)
27) A(0) B(2) C(4)
28) A(0) B(3) C(3)

added some more and we are equal - total 28 ways



[/spoiler]
yellowho wrote:In how many ways can 6 identical coins be distributed among Alex, Bea and Chad? Note: Some people may receive zero coins.


I don't have solution to this. Is this 28?
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by Night reader » Tue Mar 01, 2011 11:24 pm
I thought about this :) there's no distribution then - just 0s for everyone
rohu27 wrote:can we have a case here where a=b=c=0?
Night reader wrote:IOM [spoiler]because some of three people (ABC) may receive zero coins, which are identical
1) A(6) B(0) C(0)
2) A(5) B(1) C(0)
3) A(5) B(0) C(1)
4) A(4) B(2) C(0)
5) A(4) B(0) C(2)
6) A(4) B(1) C(1)
7) A(3) B(3) C(0)
8) A(3) B(0) C(3)
9) A(3) B(2) C(1)
10) A(3) B(1) C(2)
11) A(2) B(4) C(0)
12) A(2) B(0) C(4)
13) A(2) B(3) C(1)
14) A(2) B(2) C(2)
15) A(2) B(1) C(3)
16) A(1) B(5) (0)
17) A(1) B(0) C(5)
18) A(1) B(1) C(4)
19) A(1) B(4) C(1)
20) A(1) B(3) C(2)
21) A(1) B(2) C(3)
22) A(0) B(6) C(0)
23) A(0) B(0) C(6)
24) A(0) B(5) C(1)
25) A(0) B(1) C(5)
26) A(0) B(4) C(2)
27) A(0) B(2) C(4)
28) A(0) B(3) C(3)

added some more and we are equal - total 28 ways



[/spoiler]
yellowho wrote:In how many ways can 6 identical coins be distributed among Alex, Bea and Chad? Note: Some people may receive zero coins.


I don't have solution to this. Is this 28?
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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by anshumishra » Wed Mar 02, 2011 3:54 am
yellowho wrote:In how many ways can 6 identical coins be distributed among Alex, Bea and Chad? Note: Some people may receive zero coins.


I don't have solution to this. Is this 28?
The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items(≤ n) is : (n+r-1)C(r-1)]

So, here the number of ways = (6+3-1)C(3-1) = 8C2 = 7*8/2 = 28.
Thanks
Anshu

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