There are n identical balls and m different bins. Let Bn,m be the total number of way to put n balls in m bins.
(a) What are B1,1, B2,1, B1,2, B2,2, B3,2?
To solve the problem, we need to understand the concept of combinations and the formula for calculating combinations.
Combination refers to the number of ways to choose items from a larger set without any regard to the order of selection. The formula for calculating combinations is given by:
C(n, r) = n! / (r! * (n - r)!)
Where:
- C(n, r) represents the combination of choosing r items from a set of n items.
- n! represents the factorial of n, which is the product of all positive integers from 1 to n.
Now, let's calculate the values of Bn,m using the given values.
(a) Calculating B1,1:
In this case, we have 1 identical ball and 1 bin. Since there is only one ball and one bin, there is only one way to put the ball in the bin. Hence, B1,1 = 1.
(b) Calculating B2,1:
Here, we have 2 identical balls and 1 bin. Since there is only one bin, both balls have to be put in that bin. Therefore, there is only one way to put the balls in the bin. Hence, B2,1 = 1.
(c) Calculating B1,2:
In this case, we have 1 identical ball and 2 bins. The ball can be put in either of the two bins. Therefore, there are two ways to put the ball in the bins. Hence, B1,2 = 2.
(d) Calculating B2,2:
For 2 identical balls and 2 bins, we can have the following possibilities:
- Both balls in one bin: There is only one way of doing this.
- One ball in each bin: Again, there is only one way of doing this.
Therefore, B2,2 = 2.
(e) Calculating B3,2:
With 3 identical balls and 2 bins, we can have the following possibilities:
- All 3 balls in one bin: There is only one way of doing this.
- One bin with 2 balls and the other bin with 1 ball: Again, there is only one way of doing this.
Therefore, B3,2 = 2.
In summary:
- B1,1 = 1
- B2,1 = 1
- B1,2 = 2
- B2,2 = 2
- B3,2 = 2