In how many ways 5 diffeeent balls can be arranged into 3 different boxes so that no box remains empty? Plz explain it

when 5 diff balls are arranged into 3 diff boxes then 2 diff cases occures

either empty groups are allowed or not

but A.T.Q.

if no box remain empty then the formula used is -

n-1_Cr-1 

here n=5  & r=3

^5-1C_3-1    then by solving this

answer= 6

The answer is simple....... 5!/3!.......

First choose and 3 balls - 5C3 then arrange them into the boxes- 3! now each box has atleast 1 ball. So left over 2 balls can be arrannged into boxes in 3C2 ways 

therefore total 5C3.3!.3C2 = 180