Question

There are 5 boys and 5 girls on a co-ed basketball team. a team plays 5 players on the court at one time. how many possible ways are there to field a team of 3 boys and 2 girls? position doesn’t matter.

Answer 1

3spots for 5 boys

5(4)(3)= 60

2spots for 5girls

(5)(4)=20

60+20=80

I'm not sure if that's correct, it's been a few years

Answer 2

Explanation:

Combination is used for selection

Combination formula

nCr = n! / ( n-r)! r!

Total boys = 5

Total girls = 5

How may way can we have 3boys and 2 girls, a team comprises of 5 players

To have 3 boys means we are selecting 3 boys from the 5 boys and to have 2 girls Is that we are selecting 2 girls from the 5 girls.

3boys and two girls

Therefore, we have

5C3 × 5C2

[5!/(5-3)!3!] × [5!/(5-2)!(2!)]

[5! / (2! 3!)] × [5! / (3!2!)]

(5×4×3! / 2× 1 × 3!) × ( 5×4×3!/3! × 2 ×1)

10 × 10 = 100ways

There are 100 possible ways to field a team of 3 boys and 2 girls