A box contains $6$ balls which may be all of different colours or three each of two colours or two each of three different colours. The number of ways of selecting $3$ balls from the box (if ball of same colour are identical), is $N$ , then sum of the digits in the number ' $N$ ' equals.

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Solution

$\begin{matrix} t h e r e a r e t h r e e c o n d i t i o n s : 1) c a s e : A l l d i f f e r e n t : N o . o f s e l e c t i n g 3 b a l l s =^{6} c_{3} = \frac{6!}{3! .3!} = \frac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1 \times 3!} = 5 \times 4 = 20 2) c a s e : 3 b a l l s e a c h o f 2 c o l o u r s : (i n w h i c h 3 i s r e d, 3 i s w h i t e o r r e d, 2 w h i t e) (i) 3 r e d o r 3 w h i t e o r =^{3} c_{3} +^{3} c_{3} = 1 + 1 = 2 (i i) 2 r e d o r 1 w h i t e : =^{3} c_{2} +^{3} c_{3} \times 2 = (3 \times 3) \times 2 = 18 s o t h e t o t a l w a y s : 18 + 2 = 20 3) c a s e : 2 b a l l s e a c h o f 3 d i f f . c o l o r s : (2 r e d, 2 w h i t e, 2 b l u e) i) 1 b a l l o f e a c h c o l o r :^{3} c_{2} \times^{3} c_{2} \times^{3} c_{2} = 3 \times 3 \times 3 = 27 i i) 2 b a l l o f 1 s t c o l o r & 1 b a l l o f 2 n d c o l o r :^{2} c_{2} \times^{4} c_{1} \times 3 = 4 \times 3 = 12 s o, t h e t o t a l w a y i s 27 + 12 = 39 S o t h a t, w e h a v e t o t a l n o . o f w a y s i s : 20 + 20 + 39 = 79 \end{matrix}$

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