Q. 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) $42(4!)$
B) $2^{6} \times 4!$
C) $\left( 2^{6} - 1 \right)(4!)$
D) None of these
Explanation
(a)
Required number of selections
$\ ^{3}C_{1} \times^{4}C_{1} \times^{2}C_{1}(^{6}C_{3}{+ \ }^{6}C_{2}{+ \ }^{6}C_{1}{+ \ }^{6}C_{0})$
$= 3 \times 4 \times 2(20 + 15 + 6 + 1) = 42(4!)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.