There are twenty numbered balls in a bag.Two of the balls are numbered 0,six are numbered 1, five are numbered 2 and seven are numbered 3.

Four of these balls are chosen at random,without replacement.Calculate the different number of ways this can be done so that the four balls have that total 3.

I got it. answer is 70

To calculate the different number of ways in which four balls can be chosen at random without replacement and have a total of 3, we need to consider all possible combinations.

First, let's determine the different number of ways to pick the balls with the constraint that the total is 3. We can start by considering different cases:

Case 1: Selecting one ball with number 0 and three balls with number 3:
- Number of ways to select one ball with number 0: 2 (as there are two balls numbered 0)
- Number of ways to select three balls with number 3: (7 choose 3) = (7! / (3! * (7-3)!)) = 35

Case 2: Selecting two balls with number 0, one ball with number 2, and one ball with number 3:
- Number of ways to select two balls with number 0: (2 choose 2) = 1
- Number of ways to select one ball with number 2: (5 choose 1) = 5
- Number of ways to select one ball with number 3: (7 choose 1) = 7
- Total number of ways for this case: 1 * 5 * 7 = 35

Case 3: Selecting three balls with number 0 and one ball with number 1:
- Number of ways to select three balls with number 0: (2 choose 3) = 0 (since there are only two balls numbered 0)
- Number of ways to select one ball with number 1: (6 choose 1) = 6

Thus, the total number of ways to choose four balls with a total of 3 is:
35 + 35 + 0 = 70.

So, there are 70 different ways to select four balls from the bag such that the total of the selected balls is 3.