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Q. 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.

Permutations and Combinations

Solution:

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$
$=84+7=91$