Each bag C contains= n coins
thus, n no. of bag C contain = n × n= n²
Now n no. of bag C = 1 bag B
therefore 1 bag B contains =n² coins
If a bag B is removed from bag A, final no. of bag B in bag A = (n-1) [as there were n no. of bag B in bag A initially].
thus (n-1) no. of bag B will contain= (n-1) × n² coins
now bag A = (n-1) no. of bag B
Thus coins left in a bag
A is n²(n-1) [option 4]