In a class of 10 students there are 3 girls, A, B, C. In how many different ways can they be arranged in a row such that no two of the three girls are consecutive?

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Solution

After arranging boys in 7! ways we will have 8 places in which we can arrange the girls in $^{8} P_{3}$ ways
Hence by fundamental theorem the number of arrangements is
7! $\times^{8} P_{3} = 7! \times \frac{8!}{5!} = 7! \times 8 \times 7 \times 6 = 336 \times 7!$

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In a class of 10 students there are 3 girls, A, B, C. In how many different ways can they be arranged in a row such that no two of the three girls are consecutive?

After arranging boys in 7! ways we will have 8 places in which we can arrange the girls in 8P3 waysHence by fundamental theorem the number of arrangements is7! ×8P3=7!×8!5!=7!×8×7×6=336×7!

After arranging boys in 7! ways we will have 8 places in which we can arrange the girls in $^{8} P_{3}$ ways
Hence by fundamental theorem the number of arrangements is
7! $\times^{8} P_{3} = 7! \times \frac{8!}{5!} = 7! \times 8 \times 7 \times 6 = 336 \times 7!$