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Sagot :
To determine the number of ways to choose a committee of 6 members from a student club of 15 members, we use a concept from combinatorics called combinations. The number of combinations is given by the formula:
[tex]\[ C(n, k) = \frac{n!}{k!(n-k)!} \][/tex]
Where:
- [tex]\( n \)[/tex] is the total number of members,
- [tex]\( k \)[/tex] is the number of members to be chosen for the committee,
- [tex]\( ! \)[/tex] denotes factorial, which is the product of an integer and all the integers below it.
In this case:
[tex]\[ n = 15 \][/tex]
[tex]\[ k = 6 \][/tex]
So, we need to calculate:
[tex]\[ C(15, 6) = \frac{15!}{6!(15-6)!} = \frac{15!}{6! \cdot 9!} \][/tex]
After evaluating the above expression (not explicitly calculated here), we find that:
[tex]\[ C(15, 6) = 5005 \][/tex]
Therefore, the number of ways to choose a committee of 6 members from 15 members is:
[tex]\[ \boxed{5005} \][/tex]
The correct answer is:
O C. 5,005
[tex]\[ C(n, k) = \frac{n!}{k!(n-k)!} \][/tex]
Where:
- [tex]\( n \)[/tex] is the total number of members,
- [tex]\( k \)[/tex] is the number of members to be chosen for the committee,
- [tex]\( ! \)[/tex] denotes factorial, which is the product of an integer and all the integers below it.
In this case:
[tex]\[ n = 15 \][/tex]
[tex]\[ k = 6 \][/tex]
So, we need to calculate:
[tex]\[ C(15, 6) = \frac{15!}{6!(15-6)!} = \frac{15!}{6! \cdot 9!} \][/tex]
After evaluating the above expression (not explicitly calculated here), we find that:
[tex]\[ C(15, 6) = 5005 \][/tex]
Therefore, the number of ways to choose a committee of 6 members from 15 members is:
[tex]\[ \boxed{5005} \][/tex]
The correct answer is:
O C. 5,005
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