During the Mathematical Daily Assessment, students are asked to do 7 of the 10 questions available. Many ways students choose the number of questions to work on is.. A.115, B.120, C.125, D.130, E.135

Question

Please someone help me. I would really appreciate it.

Answer 1

To do this problem, you would need to determine if this is a permutation or a combination. In this case, because answering questions 1, 2, 3, 4, 5, 6, and 7 is the same as answering questions 7, 6, 5, 4, 3, 2, and 1, this is a combination.

There are a total of 10 possible questions to choose and each person must choose 7 of them. That means you would have ₁₀C₇

If you have a calculator, such as a TI-84, you can use the combination function that is found in "math", "PROB", then number 3 (nCr) and doing so gives you ₁₀C₇ = 120.

If you don't, you can use the following equation:

_nC_r = n!/(r!(n - r)!

Note: "!" means Factorial which means you take that number then multiply it by one less, then by one less, then etc, etc until you get to 1. Example: 7! = 7•6•5•4•3•2•1

So, using the equation with n = 10 and r = 7 we get:

₁₀C₇ = 10!/(7!(10 - 7)! = (10•9•8•7•6•5•4•3•2•1)/[(7•6•5•4•3•2•1)(3!) = (10•9•8)/(3•2•1) = 720/6 = 120

Answer 2

Choose combinations of 7 out of a set of 10. (https://en.wikipedia.org/wiki/Combination)

(10 / 7) = 10! / (7! 3!) = 10*9*8/(3.2) = 10*3*4 = 120