Bags of marbles combination/permutation problem

[Aztral]

Okie. Here goes...

Let's say you have B bags of marbles B1, B2,...Bb (all marbles are unique);
The number of marbles in each bag varies. B1 has n1 marbles, B2 has n2 marbles, etc.

You want to to create a sequences of B marbles
{marble1, marble2,...marbleb}, where marble1 came from B1, marble2 from B2, etc.

The total number of permutations is n1*n2*...nb.

But...suppose you now only allow m marbles in each permutation to be duplicated.

//////////////////Questionow many permutations are there now?//////////////////////



Ex: You have 3 bags of marbles.
B1 {red,green,blue}, n1 = 3;
B2 {white,black}, n2 = 2;
B3 {silver,gold}, n3 = 2;

I will only allow 1 marble to be duplicated, so

{red,white,silver}
{red,black,gold}
{green,white,gold}
{green,black,silver}

Can't use blue now because the options would be...
{blue,white,gold}// { x, white, gold} was used in {green,white,gold} so would have 2 duplicated marbles
{blue,white,silver}// { x, white, silver} was used in {red,white,silver} so would have 2 duplicated marbles
{blue,black,gold}// { x, black,gold} was used in {red,black,gold} so would have 2 duplicated marbles
{blue,black,silver}// { x, black, silver} was used in {green,black,silver} so would have 2 duplicated marbles



Any help would be appreciated.

Thx! :)

 

[martial]

Try to figure it out for your small example and then generalize it for your homework problem.

 

[tacman]

In this scenario, the number of permutations would be reduced because we are limiting the number of duplicated marbles. To calculate the number of permutations, we can use the formula for combinations with repetitions, which is n^r, where n is the number of options and r is the number of choices. In this case, n would be the number of marbles in each bag, and r would be the number of bags.

So for the first permutation, we have 3 options for the first marble, 2 options for the second marble, and 2 options for the third marble. This gives us a total of 3*2*2 = 12 permutations. Similarly, for the second permutation, we have 3*2*2 = 12 permutations.

However, as you mentioned, we cannot use the blue marble in any of the permutations because it would result in more than one duplicated marble. So for the remaining two permutations, we only have 2 options for the first marble, 2 options for the second marble, and 2 options for the third marble. This gives us a total of 2*2*2 = 8 permutations.

Therefore, the total number of permutations with only one duplicated marble allowed would be 12 + 12 + 8 + 8 = 40. This is significantly less than the total number of permutations without any restrictions, which was 3*2*2 = 24.

I hope this helps answer your question. Let me know if you need any further clarification. Good luck with your problem!