In how many ways can you put seven marbles in different colors into two jars? Note that the jars may be empty.

14384 ways

Step-by-step explanation:

With 0 identical marbles permitted to be included in any of the jars, An expression can be developed to determine the total of marbles in jar arrangements, which is:

E = [(n+j -1)!]*{1/[(j-1)!]*[(n)!]}, where n = number of identical balls and j =number of distinct jars, the contents of all of which must sum to n for each marbles in j jars arrangement. With n = 7 and j = 4. E = 10!/(3!)(7!) = 120= number of ways 7 identical marbles can be distributed to 4 distinct jars such that up to 3 boxes may be empty and the maximum to any box is 7 balls.

The marble arrangements are: (7,0,0,0) in 4!/3! = 4 ways, (6,1,0,0) in 4!/2! = 12 ways, (5,2,0,0) in 4!/2! = 12 ways, (5,1,1,0) in 4!/2! = 12 ways, (4,3,0,0) in 4!/2! = 12 ways, (4,2,1,0) in 4! = 24 ways, (4,1,1,1) in 4!/3! = 4 ways, (3,3,1,0) in 4!/2! = 12 ways, (3,2,2,0) in 4!/2! = 12 ways, (3,2,1,1) in 4!/2! = 12 ways, (2,2,2,1) in 4!/3! = 4 ways.

Total of ways = 4+12+12+12+12+24+4+12+12+12+4 = 120 as previously determined above for identical marbles and distinct jars.

Taking into account distinct colored marbles, the number of ways of marble distribution into 4 jars becomes as follows:

For (7,0,0,0) = 4*(7!/7!) =4. For (6,1,0,0) = 12*[7!/(6!)(1!)] = 84. For (5,2,0,0) =

12*[7!/(5!)(2!)] = 252. For (5,1,1,0) = 12*[7!/(5!)(1!)(1!)] = 504. For (4,3,0,0) =

12*[7!/(4!)(3!)] = 420. for (4,2,1,0) = 24*[7!/(4!)(2!)(1!)] = 2,520. For (4,1,1,1) =

4*7!/(4!)(1!)(1!)(1!)] = 840. For (3,3,1,0) = 12*]7!/(3!)(3!)(1!) = 1,680. For (3,2,20) = 12*]7!/(3!)(2!)(2!) = 2,520. For (3,2,1,1) = 12*]7!/(3!)(2!)(1!)(1!) = 5,040. For (2,2,2,1) = 4*]7!/(2!)(2!)(2!)(1!) = 2,520.

Total of ways as requested for distinct colored marbles and distinct jars = 4+84+252+504+420+2,520+840+1,680+2,520+5,040+2,520 = 14,384.

you can use a^2+ b^2=c^2