A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work mus be shown. If an examinee must answer 6 of the multiple-choice questions and 4 of the open-ended problems, in how many ways can the questions and problems be chosen?
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261,273,600
1260
1,296
21,772,800
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The number of ways to choose 6 multiple-choice questions out of 9 is given by the combination formula C(n, k) = n! / [k!(n-k)!], where n is the total number of items, k is the number of items to choose, and ! denotes factorial.
So, the number of ways to choose the multiple-choice questions is C(9, 6) = 9! / [6!(9-6)!] = 84.
Similarly, the number of ways to choose 4 open-ended problems out of 6 is C(6, 4) = 6! / [4!(6-4)!] = 15.
Since the choices of multiple-choice questions and open-ended problems are independent, the total number of ways to choose the questions and problems is the product of the two, which is 84 * 15 = 1260.
So, the correct answer is 1260.
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