A student has to take seven more courses before she can graduate. If none of the courses are prerequisites to others, how many groups of five courses can she select for the next semester?
In this question, "n=7" and "x=5", to find the number of groups that the student can select, the concept of combinations is applied.
The combinations formula is given as, "\\binom{n}{x}=n!\/(n-x)!*x!"
Putting the values for "n" and "x" in the formula,
number of groups "=7!\/(2!*5!)=7*6\/2*1=21"
Therefore, the number of groups of five courses the student can select for the next semester is 21.
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