Question
A bag contains 3 black, 4 white and 2 red balls, all the balls being different. The number of selections of at most 6 balls containing balls of all the colours is
A.
$$42\left( {4!} \right)$$
B.
$${{2^6} \times 4!}$$
C.
$$\left( {{2^6} - 1} \right)\left( {4!} \right)$$
D.
None of these
Answer :
$$42\left( {4!} \right)$$
Solution :
The required number of selections
$$ = {\,^3}{C_1} \times {\,^4}{C_1} \times {\,^2}{C_1}\left( {^6{C_3} + {\,^6}{C_2} + {\,^6}{C_1} + {\,^6}{C_0}} \right).$$