Question

Mr Smith has a bag A. Bag A has "n" numbers of bag B. Each bag B has " n " numbers of bag C. Each bag C has " n" number of one rupee coins in it. If one bag B is removed from bag A, then the total number of coins left in bag A is.....1) n(n-1)(n+1)2) n(n-1)3) n2 (n+1)4) n2(n-1)

Answer 1

HI!!
THIS IS YOUR ANSWER!!

4) ${n}^{2} (n - 1)$

I HOPE IT MAY HELP!!

Answer 2

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 bag A is n²(n-1) [option 4]