Is this the right way to answer this question?
A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee must answer 7 of the multiple-choice questions and 3 of the open-ended problems, in how many ways can the questions and problems be chosen?
(9x8x7x6x5x4x3)x (6x5x4) = 21772800
The number of possibilities is nCk, or n choose k.
nCk = n! / (k! (n-k)!)
For example, the number of possibilities for the multiple choice questions is 9C7.
There are 2 types of questions - multiply the number of possibilities for each question.
I tried that and it doesn't seem to add up.
n! = 9x8x7x6x5x4x3x2 / (k= 7x6x5x4x3x2(2) = 362880/5040(2)=362880/10080 = 36 there seems to be more possibilities than that
For the multiple choice questions, the examinee is choosing 7 out of 9 questions. There are not so many ways you can choose 7 of those 9 - remember order does not matter.
Also, after you compute the 6C3 for the open ended, and multiply them, you end up with 720 possibilities total.
Thank you thank you!! I understand now!
Use "Votes for Women Broadside," Women's Political Union, New York, 28 January 1911 to answer the question.
The image calls for which of the following social changes? Explain
A. women's suffrage
B. suffrage for felons
C. equal pay for women
D. suppression of women's votes
A. Women's suffrage.
The "Votes for Women Broadside" supports the idea of women's suffrage, or the right of women to vote. It uses strong, persuasive language to call for a change in society that allows women to have the same political rights as men. The text on the broadside argues that if women are good enough to be mothers, they are good enough to vote, and that women need the vote to protect themselves and their families. Overall, the image is a clear and powerful call for women's suffrage and the social change it represents.
Yes, your answer is correct.
To explain how you arrived at the answer, let's break down the problem:
We have 9 multiple-choice questions and we need to choose 7 of them. We can do this in 9 choose 7 ways, which is written as 9C7 or C(9,7). This can be calculated as:
9C7 = (9!)/(7!(9-7)!) = (9x8)/(2x1) = 36
Similarly, we have 6 open-ended problems and we need to choose 3 of them. We can do this in 6 choose 3 ways, which is written as 6C3 or C(6,3). This can be calculated as:
6C3 = (6!)/(3!(6-3)!) = (6x5)/(2x1) = 10
To find the total number of ways of choosing questions and problems, we need to multiply the number of ways of choosing the multiple-choice questions by the number of ways of choosing the open-ended problems:
Total number of ways = 9C7 x 6C3 = 36 x 10 = 360
Therefore, there are 360 ways to choose the questions and problems.