Math  /  Discrete

QuestionThe student body of 15 students wants to elect four representatives. How many possibilities are there?\text{The student body of 15 students wants to elect four representatives. How many possibilities are there?}

Studdy Solution
Calculate the number of combinations:
C(15,4)=15!4!×11! C(15, 4) = \frac{15!}{4! \times 11!}
Calculate the factorials:
15!=15×14×13×12×11! 15! = 15 \times 14 \times 13 \times 12 \times 11!
4!=4×3×2×1=24 4! = 4 \times 3 \times 2 \times 1 = 24
11!=11! 11! = 11!
Substitute back into the equation:
C(15,4)=15×14×13×12×11!4!×11! C(15, 4) = \frac{15 \times 14 \times 13 \times 12 \times 11!}{4! \times 11!}
Cancel out 11! 11! :
C(15,4)=15×14×13×1224 C(15, 4) = \frac{15 \times 14 \times 13 \times 12}{24}
Calculate the result:
C(15,4)=3276024=1365 C(15, 4) = \frac{32760}{24} = 1365
The number of possibilities to elect four representatives is:
1365 \boxed{1365}

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