Question

From a class of 12 boys and 10 girls, 10 students are to be chosen for a competition; at least including 4 boys and 4 girls. The 2 girls who won the prizes last year should be included. In how many ways can the selection be made ?

Answer 1

Total number of boys = 12

Total number of girls = 10

Total number of girls for the competition = 10 + 2 = 12

Total students chosen for competition

= 10 - 2(at least 4 boys and 4 girls)

$\therefore$ Selection can be made in

${}^{12}{C}_4{\times }^8{C}_4{\times }^{12}{C}_5{\times }^8{C}_3{\times }^{12}{C}_6{\times }^8{C}_2$

$=\frac{12!}{4!8!}\times \frac{8!}{4!4!}+\frac{12!}{5!7!}\times \frac{8!}{3!5!}+\frac{12!}{6!6!}\times \frac{8!}{2!6!}$

$=\left(\frac{12\times 11\times 10\times 9\times 8\times 7\times 6\times 5}{4\times 3\times 2\times 4\times 3\times 2}\right)$

$+\left(\frac{12\times 11\times 10\times 9\times 8\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 3\times 2}\right)$

$+\left(\frac{12\times 11\times 10\times 9\times 8\times 7\times 8\times 7}{6\times 5\times 4\times 3\times 2\times 2}\right)$

$=55440+44352+181104$

$=280896$

$\therefore$ Total number of ways

= 385770 - 280896 = 104874

(385770 = from 10 girls 4 are chosen)