Question

A group consists of 4 girls and 7 boys.
Statement I The number of ways a team of 5 members can be selected, if the team has no girls are 21.
Statement II The number of ways a team of 5 members can be selected, if it has atleast one boy and one girl are 441.
Statement III The number of ways a team of 5 members can be selected, if it has atleast 3 girls are 90.
Identify the correct combination of true (T) and false $(F)$ of the given three statements.

• A. FFT
• B. TFF
• C. TTF
• D. FTT

Answer 1

Answer: (C)

I. Since, the team will not include any girl, therefore, only boys are to be selected. 5 boys out of 7 boys can be selected in ${}^7{C}_5$ ways. Therefore, the required number of ways $={}^7{C}_5=\frac{7!}{5!2!}=\frac{6\times 7}{2}=21$
II. Since, atleast one boy and one girl are to be there in every team. Therefore, the team can consist of
(a) 1 boy and 4 girls
(b) 2 boys and 3 girls
(c) 3 boys and 2 girls
(d) 4 boys and 1 girl.
1 boys and 4 girls can be selected in ${}^7{C}_1\times {}^4{C}_4$ ways. 2 boys and 3 girls can be selected in ${}^7{C}_2\times {}^4{C}_3$ ways.
3 boys and 2 girls can be selected in ${}^7{C}_3\times {}^4{C}_2$ ways.
4 boys and 1 girl can be selected in ${}^7{C}_4\times {}^4{C}_1$ ways.
Therefore, the required number of ways
$={}^7{C}_1\times {}^4{C}_4+{}^7{C}_2\times {}^4{C}_3+{}^7{C}_3\times {}^4{C}_2+{}^7{C}_4\times {}^4{C}_1$
$=7+84+210+140=441$
III. Since, the team has to consist of atleast 3 girls, the team can consist of
(a) 3 girls and 2 boys, or
(b) 4 girls and 1 boy.
Note that the team cannot have all 5 girls, because the group has only 4 girls.
3 girls and 2 boys can be selected in ${}^4{C}_3\times {}^7{C}_2$ ways.
4 girls and 1 boy can be selected in ${}^4{C}_4\times {}^7{C}_1$ ways.
Therefore, the required number of ways
$={}^4{C}_3\times {}^7{C}_2+{}^4{C}_4\times {}^7{C}_1$
$=84+7=91$