# Moderate Permutation-Combination Solved QuestionAptitude Discussion

 Q. A $6\times 6$ grid is cut from an $8\times 8$ chessboard. In how many ways can we put two identical coins, one on the black square and one on a white square on the grid, such that they are not placed in the same row or in the same column?
 ✔ A. 216 ✖ B. 324 ✖ C. 144 ✖ D. 108

Solution:
Option(A) is correct

In a $6\times6$ grid of a chessboard, each row and each column contains $3$ white and $3$ black squares placed alternatively.

There are a total of $18$ black and $18$ white squares.

For every black square chosen to put one coin, we cannot choose any white square present in its row or column.

There are $3$ white squares in its row and $3$ white square in its column for every black square.

Hence for every black square chosen, we can choose $(18-6) =12$ white squares.

Total number of possibilities where a black square and a white square can be chosen so that they do not fall in the same row or in the same column:

$= 18×12 = 216$

So there are $\textbf{216}$ ways of placing the coins that are identical.

## (2) Comment(s)

PRATYUSH ANAND
()

In 6x6 grid ,there will be 36 positions or spots.In which 18 White and 18 Black spots. Thes spots are Black and White alternatively like Chess Board.

So selecting one position out of 18 White spots for placing 1st coin =18C1

As 2nd coin must not be placed --

(i)in same row and column.

& (ii) in same color spots where 1st coin placed.

So, no. of Black spots, where we can place our 2nd coin is (18 -6(i.e come under the same row and column of 1st coin placed))=12

So selecting one position out of 12 Black spots for placing 2nd coin= 12C1

Hence total no. of ways are 18C!x12C1=216

PRATYUSH ANAND
()

Really Good question, liked it.