Finding possible combinations

Question

The 6-member debate team plans to send 3 of its members to a conference. How many combinations of 3 members are possible from the 6 member team?

Answer 1

To determine the number of ways to select 3 members from a 6-member debate team, we use the combination formula:

Combinations=n!r!(n−r)!\text{Combinations} = \frac{n!}{r!(n-r)!}Combinations=r!(n−r)!n!​

where:

1. n=6n = 6n=6 (total members)
2. r=3r = 3r=3 (members selected)

Combinations=6!3!(6−3)!=6!3!3!\text{Combinations} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!}Combinations=3!(6−3)!6!​=3!3!6!​

Expanding the factorials:

=6×5×4×3!3!×3×2×1= \frac{6 \times 5 \times 4 \times 3!}{3! \times 3 \times 2 \times 1}=3!×3×2×16×5×4×3!​

Cancel out the 3!3!3! from numerator and denominator:

=6×5×43×2×1=1206=20= \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = \frac{120}{6} = 20=3×2×16×5×4​=6120​=20

So, there are 20 different ways to choose 3 members. ✅

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Answer 2

To start with this question, I like to think of the basic counting principle. For two or more events, the total number of outcomes is calculated by multiplying the outcomes for each event.

For your question, there are 6 choices for the first seat, 5 choices for the second seat, and 4 choices for the third. This makes 6*5*4 = 120 outcomes -- if order matters. This means that selecting students {A,B,C} is not the same as selecting students {C,B,A} or {B,C,A}. Even though it's the same collection of students, it's counted more than once. To find the true number of outcomes, we must divide by (3!) which removes the extra copies of triples. This leaves us with 120/(3!) = 20 total outcomes or combinations.

Answer 3

6!/3!3! 6(5)(4)3!/3!3! cancel 3! from top and bottom. 120/6 = 20

There are 20 combinations. Formula N!/R(N-R)!