Question 584791: A student planning her curriculum for the upcoming year must select one of three business courses, one of five mathematics courses, two of seven elective courses, and either one of four history courses or one of three social science courses. How many different curricula are available for her consideration?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! she must select:
1 of 3 business courses
1 of 5 mathematics courses
2 of 7 elective courses
and:
either 1 of 4 history courses or 1 of 3 social science courses.
using combination formulas, the possible number of choices are:
3 for the business courses
5 for the math courses
21 for the elective courses
and either:
4 for the history courses
or:
3 for the social science courses.
her possible choices will be:
3 * 5 * 21 * 4 if she chooses to do histor5y
or:
3 * 5 * 21 * 3 if she chooses to do social science4
this becomes:
1260 or 945
add these possible choices up and you get 2205 possible choices.
is this correct?
good question.
to simplify the process, assume only 2 out of 3 possible choices for business and then either 1 out 1 for history or 1 out of 1 for social science.
let b1, b2, b3 be the possible choices from business.
let h1 be the possible choice for history
let s1 be the possible choice for social science.
total number of possible choices, using the formulas in the big problem would make the number of possible selections equal to 3C2 * 1 or 3C2 * 1 which would be equal to 3 or 3 which would be equal to 6 total possible choices.
they would be:
b1 + b2 + h1
b1 + b2 + s1
b1 + b3 + h1
b1 + b3 + s1
b2 + b3 + h1
b2 + b3 + s1
that's 6 total.
now suppose she had 2 choice for history and only 1 choice for social science.
posible choices would be 3*2 + 3*1 which would be equal to 9.
those 9 possible choices would be:
b1 + b2 + h1
b1 + b2 + h2
b1 + b2 + s1
b1 + b3 + h1
b1 + b3 + h2
b1 + b3 + s1
b2 + b3 + h1
b2 + b2 + h2
b2 + b3 + s1
same formulas were used in the big problem to the assumption is that the big problem was done correctly.
at least i'm reasonably sure it was done correctly based on my assumptions of the nature of the problem.
i'm not 100% sure, but i'm reasonably sure.
combination formula used is nCx = n! / (x! * (n-x)!)
for 2 out of 7, this formula results in 7! / (2! * 5!) which results in
(7*6*5!) / (2!*5!) which becomes (7*6)/2 which becomes 42/2 which becomes 21.
the OR part is what took a little thinking.
she could do all the others plus history or she could do all the others plus social science.
this resulted in 2 separate branches that were additive.
the possible choices for the elective courses, assuming that they were e1 through e7, would be:
e1, e2
e1, e3
e1, e4
e1, e5
e1, e6
e1, e7
e2, e3
e2, e4
e2, e5
e2, e6
e2, e7
e3, e4
e3, e5
e3, e6
e3, e7
e4, e5
e4, e6
e4, e7
e5, e6
e5, e7
e6, e7
that's a total of 21 possible combinations as given by the formula.
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