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OpenStudy (anonymous):

The freshman class has 160 students. Each freshman must write three reports from a list of 12 novels that the teacher has assigned them to read. How many different combinations of novels can a freshman student choose to write reports on?

OpenStudy (anonymous):

the 160 part has nothing to do with the problem, so don't let it confuse you

OpenStudy (anonymous):

the question is this : " how may ways can 3 books be chosen out of a total of 12" or in plain math "how many ways can you choose 3 out of a set of 12" or even more briefly "what is 12 choose 3?" sometimes written as \(_{12}C_3\) or \(\binom{12}{3}\)

OpenStudy (anonymous):

do you know how to compute it?

OpenStudy (anonymous):

no never taught that

OpenStudy (anonymous):

really? then how are you expected to do it?

OpenStudy (anonymous):

I don't know

OpenStudy (anonymous):

i work off a computer

OpenStudy (anonymous):

\[_{12}C_3\] make a fraction in the top put multiply 3 numbers starting with 12 and decreasing by 1 each time, so the numerator is \(12\times 11\times 10\)

OpenStudy (anonymous):

then the denominator do the same thing, but this time starting at 3, so the denominator is \(3\times 2\times 1\)

OpenStudy (anonymous):

so \[_{12}C_3=\frac{12\times 11\times 10}{3\times 2\times 1}\] you don't need to write the 1 in the denominator

OpenStudy (anonymous):

cancel first, multiply last get \[_{12}C_3=\frac{12\times 11\times 10}{3\times 2\times 1}=2\times 11\times 10=220\]

OpenStudy (anonymous):

ok

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