Assume that you have 5 different types of vegetables and 10 different types of fruit. a vegetable salad consists of a mixture of any 3 types of vegetables, and a fruit salad consists of a mixture of 2 kinds of fruit. (1) in how many different ways can you prepare a vegetable salad and a fruit salad?

10 different vegetable salads 45 different fruit salads. The formula for picking m of n items is n!/(m! (n-m)!). To show how this formula works, let's deal with the vegetable salad problem. This problem involves using factorials. N factorial is simply 1 times 2 times 3 times ... N. This represents the number of ways you can arrange N objects and is written as N!. So imagine for the vegetable salad, you place on the table all 5 different vegetables and select the 1st 3 of the arrangement. For 5 vegetables, you can arrange them in 5! ways which is 1 * 2 * 3 * 4 * 5 = 120 different arrangements. You pick the 1st 3 vegetables and you're left with 2 vegetables you haven't selected. There are 2! ways for those leftover vegetables to ordered which is 2! = 1 * 2 = 2 and you really don't care what their order is, so divide the 120 by 2 giving you 60 different ways that you could have picked up the 3 vegetables you are going to use. But does it really matter about the order you picked up those 3 vegetables? No it doesn't. So divide once again by 3! giving you 60 / 6 = 10 different possible vegetable salads. If we had plugged in the numbers we get 5!/(3!(5-3)!) = 120/(6(2)!) = 120/(6*2) = 120/12 = 10. Now for the fruit salad, we can plug in the numbers getting 10!/(2!(10-2)!) = 3628800 / (2(8)!) = 3628800/(2*40320) = 3628800/80640 = 45 different fruit salads

It is 6% because if you subtract 69-58=6 so 6% is the answer to your question.