Question

A group consists of $4$ girls and $7$ boys. In how many ways can a team of $5$ members be selected if the team has (i) no girls (ii) atleast one boy and one girl (iii) atleast three girls?

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (C)

c

Answer 2

Number of girls $= 4$, Number of boys $= 7$ We have to select a team of $5$ members if the team (i) having no girl $\therefore$ Required no. of ways = $^{7}C_{5}$ $= \frac{7\times 6 }{2}$ $= 21$ (ii) having atleast one boy and one girl $\therefore$ Required no. of ways $= (^{7}C_{1} \times\, ^{4}C_{4}) +\, (^{7}C_{2} \times\,^ {4}C_{2}) +\, (^{7}C_{3} \times\, ^{4}C_{2}) +\, (^{7}C_{4} \times\, ^{4}C_{1})$ $= 7 + 84 + 210+ 140$ $= 441$ (iii) having atleast three girls $\therefore$ Required no. of ways $= (^{4}C_{3} \times\, ^{7}C_{2}) + \,(^{4}C_{4} \times \,^{7}C_{1})$ $= 91$