Question

A box contains three white balls, four black balls, and three red balls. The number of ways in
which three balls can be drawn from the box so that all the three drawn balls will have different
colours is?​

Answer 1

Given :  No. of white balls = 3

             No. of black balls = 4

             No. of red balls = 3

             Total no. of balls = 10

To find : - P (all three different balls) = ?

total no. of possible selections = $n_{C_{r} } } = 10_{C_{3 } } }$

No. of ways to select three different coloured balls = $3_{C_{1} } *4_{C_{1} } *3_{C_{1} }$

∴ $P( all \ three \ different \ balls) = \frac{3_{C_{1} } *4_{C_{1} } *3_{C_{1} }}{10_{C_{3} } }$

⇒  P (all three different balls) = $\frac{(3*4*3)}{(\frac{10*9*8}{3*2*1} )}}$

∴  P (all three different balls) = $\frac{3}{10}$

HOPE THIS HELPS YOU .!! ✌️☘️

Answer 2

Step-by-step explanation:

Given : No. of white balls = 3

No. of black balls = 4

No. of red balls = 3

Total no. of balls = 10

To find : - P (all three different balls) = ?

total no. of possible selections = n_{C_{r} } } = 10_{C_{3 } } }

No. of ways to select three different coloured balls = 3_{C_{1} } *4_{C_{1} } *3_{C_{1} }3

C

1

∗4

C

1

∗3

C

1

∴ P( all \ three \ different \ balls) = \frac{3_{C_{1} } *4_{C_{1} } *3_{C_{1} }}{10_{C_{3} } }P(all three different balls)=

10

C

3

3

C

1

∗4

C

1

∗3

C

1

⇒ P (all three different balls) = \frac{(3*4*3)}{(\frac{10*9*8}{3*2*1} )}}

∴ P (all three different balls) = \frac{3}{10}

10

3

HOPE THIS HELPS YOU .!! ✌️☘️