Can you mention the source of the question please

Can you mention the source of the question please

no

no

Hi Aaditya. Since we have a 5-faced unbiased dice, we can say that the numbers on the dice faces must be 1,2,3,4,5. Now, we are getting a sum of 14 after the first 4 throws, and at least one of them is a 3.So, the sum of outcome of other 3 throws is 11.So, the possible cases are:(3,5,5,1) -> 4! / 2! = 12 cases(3,5,4,2) -> 4! = 24 cases(3,5,3,3) -> 4! / 3! = 4 cases(3,4,4,3) -> 4! / 2!^2 = 6 casesSo, the total cases are 46.

Hi Aaditya. Since we have a 5-faced unbiased dice, we can say that the numbers on the dice faces must be 1,2,3,4,5. Now, we are getting a sum of 14 after the first 4 throws, and at least one of them is a 3.So, the sum of outcome of other 3 throws is 11.So, the possible cases are:(3,5,5,1) -> 4! / 2! = 12 cases(3,5,4,2) -> 4! = 24 cases(3,5,3,3) -> 4! / 3! = 4 cases(3,4,4,3) -> 4! / 2!^2 = 6 casesSo, the total cases are 46.

give vid n audio solution Rajesh rolled a 5 faced unbiased dice continuously and observed that the sum of the numbers obtained after the first 4 throws was 14. How many unique combinations of throws are possible such that at least one of the throws showed 3?

give vid n audio solution Rajesh rolled a 5 faced unbiased dice continuously and observed that the sum of the numbers obtained after the first 4 throws was 14. How many unique combinations of throws are possible such that at least one of the throws showed 3?