Three dice (each having six faces with each face having one number from 1 to 6) are rolled.

(c) There can be 3 cases : 

I. When one dice shows 2. 

II. When two dice shows 2. 

III. When three dices shows 2. 

Case I : The dice which shows 2 can be selected out of the 3 dices in ³C₁ ways.

Remaining 2 dices can have any 5 numbers except 2. So number of ways for them = ⁵C₁ each, so no of ways when one dice shows 2 = ³C₁ × ⁵C₁ × ⁵C₁. 

Case II : Two dices, showing 2 can be selected out of the 3 dices in ³C₂ ways and the rest one can have any 5 numbers except 2, so number of ways for the remaining 1 dice = 5. 

So, number of ways, when two dices show 2 = ³C₂ × 5 

Case III : When three dices show 2 then these can be selected in ³C₃ ways.

So, number of ways, when three dices show 2 = ³C₃ = 1 

As, either of these three cases are possible. 

Hence, total number of ways 

= (3 × 5 × 5) + (3 × 5) + 1 = 91