A team consists of 6 boys and 4 girls and other has 5 boys and 3 girls. How many single matches can be arranged between the two teams when a boy plays against a boy and a girl plays against a girl?

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Solution

A boy can be selected from the first team in 6 ways, and from the second in 5 ways, So, number of single matches between the boys of two teams = $6 \times 5 = 30$ Similarly, the number of single matches between the girls of two teams = 4 \times 3 = 12 So, total number of matches = 30+12= 42.

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A team consists of 6 boys and 4 girls and other has 5 boys and 3 girls. How many single matches can be arranged between the two teams when a boy plays against a boy and a girl plays against a girl?

A boy can be selected from the first team in 6 ways, and from the second in 5 ways, So, number of single matches between the boys of two teams = 6×5=30 Similarly, the number of single matches between the girls of two teams = 4 \times 3 = 12 So, total number of matches = 30+12= 42.

A boy can be selected from the first team in 6 ways, and from the second in 5 ways, So, number of single matches between the boys of two teams = $6 \times 5 = 30$ Similarly, the number of single matches between the girls of two teams = 4 \times 3 = 12 So, total number of matches = 30+12= 42.