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Q. The probability of getting $10$ in a single throw of three fair dice is :

BITSATBITSAT 2018

Solution:

Exhaustive no. of cases $=6^{3}$
$10$ can appear on three dice either as distinct number as following
$(1,3,6) ;(1,4,5) ;(2,3,5)$ and each can occur in $3 !$ ways.
Or $10$ can appear on three dice as repeated digits as following
$(2,2,6),(2,4,4),(3,3,4)$ and each can occur in $\frac{3 !}{2 !}$ ways.
$\therefore $ No. of favourable cases $=3 \times 3 !+3 \times \frac{3 !}{2 !}=27$