Question

You are asked to arrange the faces of a dice such that 1 and 6, 2 and 5, 3 and 4 must be on opposite faces. What is the number of ways in which you can do this?

Answer 1

Answer:

48

Explanation:

There are 3 combinations, (1,2), (3,4), (5,6).

For first three combinations i,e, for(1,2), (3,4), (5,6) the possibilities are 3!(3*2*1)=6.

For (1,2) the possibilities are 2!(2*1)=2.

For (3,4) the possibilities are 2!(2*1)=2.

For (5,6) the possibilities are 2!(2*1)=2.

∴ Total=   (3!*2!*2!*2!)=48.

Answer 2

You are asked to arrange the faces of a dice such that 1 and 6, 2 and 5, 3 and 4 must be on opposite faces. What is the number of ways in which you can do this?

since on the dice total number are 1,2.3,4,5,6.

Hence the possible combination will be 3 that are , (1,2), (3,4), (5,6).

The probability for first three combinations i,e, for(1,2), (3,4), (5,6) the possibilities are 3!

hence 3*2*1 = 6.

For (1,2) the possibilities are 2!

therefore  2*1 =2.

For (3,4) the possibilities are 2!

thus   2*1 = 2.

For (5,6) the possibilities are 2!

that is  2*1 = 2.

∴ Total will be

  (3!*2!*2!*2!) = 48.

So the final number of ways are 48 by which we can do this.