Apair of dice is tossed 10 times. what is the possibility that no 7s or 11s appear as the sum of the sides facing up?

It should be 0.08

Explanation: There are six ways to get a 7 and two ways to get an 11 on two dice, and so there are 36 – 8 = 28 ways to get anything else. Therefore, P(no 7 or 11) = P(always getting something else) = (28/36)10

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

hello i am doing good for you that it's summer in your country or state ; )

btw in my country we just have sun and rain as (because i live in the caribbean lol)

also, thx a lot for the points : )