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.....

Answer 1

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Answer 2

Answer: The total number of coins left in bag A is defined by $n^2(n-1)$.

Explanation:

It is given that 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.

We have n coin is bag C and we have n number bag C in bag B. So, number of coins in single bag B is,

$n\times n=n^2$

In bag A there are n number of bag B, So, the number of coins in bag A is,

$n\times n\times n=n^3$

The total number of coins in bag A is $n^3$.

Now, one bag B is removed from bag A.

So, now in bag A there are n-1 number of bag B, So, the number of coins in bag A is,

$n\times (n-1)\times n=n^2(n-1)$

therefore, the  total number of coins left in bag A is defined by $n^2(n-1)$.