In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
∴ Required number of ways
$$ = \left( {^6{C_1}{ \times ^4}{C_3}} \right) + \left( {^6{C_2}{ \times ^4}{C_2}} \right) + $$      $$\left( {^6{C_3}{ \times ^4}{C_1}} \right) + $$   $$\left( {^6{C_4}} \right)$$
$$ = \left( {^6{C_1}{ \times ^4}{C_1}} \right) + \left( {^6{C_2}{ \times ^4}{C_2}} \right) + $$      $$\left( {^6{C_3}{ \times ^4}{C_1}} \right) + $$   $$\left( {^6{C_2}} \right)$$
$$ = \left( {6 \times 4} \right) + \left( {\frac{{6 \times 5}}{{2 \times 1}} \times \frac{{4 \times 3}}{{2 \times 1}}} \right) + $$       $$\left( {\frac{{6 \times 5 \times 4}}{{3 \times 2 \times 1}} \times 4} \right) + $$    $$\left( {\frac{{6 \times 5}}{{2 \times 1}}} \right)$$
$$\eqalign{ & = \left( {24 + 90 + 80 + 15} \right) \cr & = 209 \cr} $$