A coin is tossed 3 times and the outcomes are recorded . how many possible outcomes are there?

Answer

7

Explanation

If `n_(1),n_(2),n_(3),n_(4)` take minimum values 1,2,3,4 respectively, then `n_(5)` will be maximum 10.
`therefore`corresponding to `n_(5)=10` there is only one solution
`n_(1)=1,n_(2)=2,n_(3)=3,n_(4)=4`
Corresponnding to `n_(5)=9`, we can have,
`n_(1)=1,n_(2)=2,n_(3)=3,n_(4)=5` i.e., one solution
Corresponding to `n_(5)=8`, we have,
`n_(1)=1,n_(2)=2,n_(3)=3,n_(4)=6`
or `n_(1)=1,n_(2)=2,n_(3)=4,n_(4)=5` i.e., two solutions
corresponding to `n_(5)=5`, we can have
`n_(1)=1,n_(2)=2,n_(3)=4, n_(4)=6`
or `n_(1)=1,n_(2)=3,n_(3)=4,n_(4)=5` i.e., two solutions
Corresponding to `n_(5)=6` we can have
`n_(1)=2,n_(2)=3,n_(3)=4,n_(4)=5` i.e., one solution
thus, there can be 7 solutions.