Another way to think about it:
Dice 1: 6 available options
And
Dice 2: 6 available options
And
Dice 3: 6 available options
We have 216 ways in which the numbers can show up on (die #1) and (die #2) and (die #3)
These 3 throws are 3 distinct throws and it will matter in the number of arrangements if we threw:
1-1-2
Vs
1-2-1
Vs
2-1-1
These 3 arrangements, for instance, are included in the 216 total arrangements with no constraints:
Case 1: 2 of new thrown dice show the same number and the 3rd die is different
1st: we can choose which of the 6 numbers will be the repeated number: can do that in 6 c 1 = 6 ways
And
2nd: the other die has to have a different number from whatever number shows up on the other two dice: 5 other numbers = 5 ways
(6) (5) = 30 ways
But: there are three scenarios in which 2 dice will show the same number and 1 of the die will show a different number. We could have:
A —— A —— B ……. (First 2 thrown are same)
A ——- B ——A ……… (first and third are same)
Or
B —— A ——-A ………. (Second and third are same)
For each one of these possibilities, there is the same 30 ways as calculated above.
(3) (30) = 90 ways
Case 2: all three dice show the same number
This is the easier case. We just have to choose which number out of 6 will appear on the three dice.
6 c 1 = 6 ways
(case 1) + (case 2) = 90 + 6 =
96 ways
Harshita6789 wrote:
In how many ways can we roll a die thrice such that at least two throws are the same?
(A) 120 (B) 51 (C) 96 (D) 66 (E) 90
Posted from my mobile device