Given five different yellow balls, four different black balls and three different white balls. Then the number of combinations of balls that can be chosen taking at least one yellow and at least one black balls are ___________.Choices:- 3255 4096 3720 3721

To solve this problem, we need to use the concept of combinations in mathematics.

First, let's calculate the total number of ways we can select balls without any restrictions. We have 5 yellow balls, 4 black balls, and 3 white balls, which gives us a total of 12 balls. The number of ways to select balls from these 12 balls is 2^12 (as each ball can be either selected or not selected), which equals 4096.

However, this total includes the cases where no yellow balls are selected and no black balls are selected, which we need to exclude.

The number of ways to select balls without any yellow balls is 2^(12-5) = 2^7 = 128 (as we have 7 balls left to choose from: 4 black and 3 white).

Similarly, the number of ways to select balls without any black balls is 2^(12-4) = 2^8 = 256 (as we have 8 balls left to choose from: 5 yellow and 3 white).

But if we subtract these two cases from the total, we are double subtracting the case where no yellow and no black balls are selected. This case is simply the number of ways to select from the 3 white balls, which is 2^3 = 8.

So, the total number of ways to select at least one yellow and at least one black ball is 4096 (total ways) - 128 (ways without yellow) - 256 (ways without black) + 8 (add back the double subtracted case) = 3720.

So, the answer is 3720.