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Home » 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) at least one boy and one girl
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) at least one boy and one girl
CBSE | Class 11 | Maths | NCERT Exemplar | Permutations and Combinations
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) at least one boy and one girl

Solution:

It is known that

{ }^{n} C_{r} =\frac{n !}{r !(n-r) !}

(i) No girls

The total no. of ways the team can have no girls ={ }^{4} \mathrm{C}_{0}{ }^{7} \mathrm{C}_{5}=21

(ii) at least one boy and one girl

Case (A) one boy and four girls

={ }^{7} C_{1} \cdot{ }^{4} C_{4} =7

Case (B) two boys and three girls

\begin{array}{l} ={ }^{7} \mathrm{C}_{2}{ }^{4} \mathrm{C}_{3} \\ =84 \end{array}

Case (C) three boys and two girls

={ }^{7} \mathrm{C}_{3} \cdot{ }^{4} \mathrm{C}_{2}

=210

Case (D) four boys and one girls

={ }^{7} \mathrm{C}_{4}{ }^{4} \mathrm{C}_{1}

=140

The total no. of ways in which the team can have at least one boy and one girl,

\begin{array}{l} =\operatorname{Case}(\mathrm{A})+\operatorname{Case}(\mathrm{B})+\operatorname{Case}(\mathrm{C})+\operatorname{Case}(\mathrm{D}) \\ =7+84+210+140 \\ =441 \end{array}

i

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