In 15 ways the parents could have 2 boys and 4 girls as children
Step-by-step explanation:
Let us revise the rule of combination (nCr)
nCr = [tex]\frac{n!}{r!(n-r)!}[/tex] , where
∵ The family has 6 children
∴ n = 6
- We need to find in how many ways the parents could have
2 boys and 4 girls
∴ r = 2 OR r = 4 ⇒ both give the same answer
∴ The number of ways is 6C2 OR 6C4
∵ 6C2 = [tex]\frac{6!}{2!(6-2)!}[/tex]
∴ 6C2 = [tex]\frac{6!}{2!(4)!}[/tex]
∵ 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
∵ 2! = 2 × 1 = 2
∵ 4! = 4 × 3 × 2 × 1 = 24
- Substitute them in 6C2
∴ 6C2 = [tex]\frac{720}{2(24)}[/tex]
∴ 6C2 = 15
∴ There are 15 ways the parents could have 2 boys and 4 girls as children
In 15 ways the parents could have 2 boys and 4 girls as children
Learn more:
You can learn more about the combination in brainly.com/question/9621364
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