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. $42\times 4!$
• B. $2^6\times 4!$
• C. $(2^6-1)(4!)$
• D. $Noneofthese$

Answer 1

The correct option is A $42\times 4!$

Since each selection should have all the three colours, minimum 3 balls one each colour should be selected in ${}^3{C}_1{\times }^4{C}_1{\times }^2{C}_1$ ways

Maximum 6 balls are to be selected

$\therefore$ the required number of ways${=}^3{C}_1{\times }^4{C}_1{\times }^2{C}_1{(}^6{C}_0{+}^6{C}_1{+}^6{C}_2+6C_3)$

$=24(1+6+15+20)$

$=42\times 4!$