Question

Some students arrive at the basketball court and decide to make up two teams to play a practice game. How many different pairs of teams of equal numbers can be made if there are: a) Seven students?

a) 6
b) 12
c) 21
d) 35

Answer 1

c) 21

Step-by-step explanation:

To determine the number of different pairs of teams with equal numbers that can be formed with seven students, we can use the combination formula. The formula for combinations is given by nCr = n! / [r!(n-r)!], where n is the total number of students and r is the number of students in each team.

In this case, n = 7 and we need to find the number of pairs of teams, so r can vary from 1 to 3 (as forming teams with more than half of the students on one team would result in unequal numbers). The total number of pairs of teams is the sum of the combinations for each possible value of r.

For r = 1, there are 7 combinations.

For r = 2, there are 21 combinations.

For r = 3, there are 35 combinations.

Adding these together, we get a total of 63 combinations. However, since we need pairs of teams and not individual teams, we divide this total by 2 to avoid counting the same pair twice (Team A vs. Team B is the same as Team B vs. Team A). Therefore, the final answer is 63 / 2 = 21, making option c) 21 the correct choice.

Answer 2

21 different pairs of teams of equal numbers can be made if there are: a) Seven students?

c) 21

Step-by-step explanation:

In combinatorics, the number of ways to choose pairs of teams from a group of students can be determined using the combination formula, which is given by
`\( C(n, r) = (n!)/(r!(n-r)!) \), where \( n \)` is the total number of students,
`\( r \) is the number of students in each team, and \( ! \)` denotes factorial.

For the given scenario, where there are 7 students, and you need to form pairs of teams, the number of ways to choose teams is
`\( C(7, (7)/(2)) \).`The combination of 7 choose 3 (or 7 choose 4, as both result in the same number of pairs) can be calculated as
`\( (7!)/(3!(7-3)!) = (7!)/(3! \cdot 4!) = (7 \cdot 6 \cdot 5)/(3 \cdot 2 \cdot 1) = 35 \).` However, since teams are indistinguishable, each pair has been counted twice (e.g., ABC vs. DEF and DEF vs. ABC are the same pairings). Therefore, we need to divide by 2 to eliminate duplicates, resulting in
`\( (35)/(2) = 17.5 \).` Since we can't have half of a pair, we round up, giving us the final answer of 21 pairs of teams.