In how many ways can we roll a die thrice such

Since at least 2 throws must be same then for each of six numbers, 1 to 6 we can have two combinations of all same or two same and one different, such as 11x,where x ranges from 1 to 6. Of these two combinations from each of the six numbers, there are 15 + 1 combinations each since 11x when x not =1 can have three ways of throw each while only one way of throw when x equals to 1. Now these 16×6=96 ways are available that satisfy the given condition. Hence, correct answer is C.

Posted from my mobile device

In atleast types of question one should go with approach: total available option - not alteast 2 means all different nos. In a row
So total available option = 6×6×6= 216
and all different nos = 6×5×4 = 120
So 216-120 = 96 is the answer

Sent from my SM-G935F using Tapatalk

(No of ways in which all 3 have different numbers) + (No of ways in which 2 are the same and 1 is different) + (No of ways in which all 3 are the same) = Total ways to throw 3 Dice


Thus


No. of ways to throw at least two side of the same number =

(Total number of ways to throw 3 dice)

-

(No. of ways to throw the dice and have all different numbers)



Total number of ways to throw 3 dice:

Dice 1——— 1 of 6 Numbers ——— 6 options


Dice 2———-1 of 6 Numbers ———6 options


Dice 3——-1 of 6 Numbers——-6 options

6 * 6 * 6 = (6)^3 = 216


No. of ways in which all three numbers are different:


Dice 1: 1 or 2 or 3 or 4 or 5 or 6 —— 6 options

Dice 2: can have any number EXCEPT the number that showed up on dice 1 ———5 options

Dice 3: 4 options

6 * 5 * 4 = 120

(216) - (120) =

96

C

Posted from my mobile device

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


Posted from my mobile device

No. of throws such that at least two throws are the same = Total number of throws - No. of throws when all throws are unique
= 6x6x6 - 6x5x4 = 216 - 120 = 96

Asked: In how many ways can we roll a die thrice such that at least two throws are the same?

Total number of ways to roll a die thrice = 6*6*6 = 216
Number of ways to roll a die thrice such that none of the throws are the same = 6*5*4 = 120
Number of ways to roll a die thrice such that at least two throws are the same = 216 - 120 = 96

IMO C