Problem Solving

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?

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?

[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]

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?

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?

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.