12 identical coins are placed in the cells on a chess board with 4 rows and 4 columns. If coins are not allowed to be placed in the same cell, how many ways are there so that there are 3 coins in every rows
and columns?

Step-by-step explanation:

If the coins are identical and not numbered, we need to find the number of patterns.

There are 4 rows and 4 columns. One cell remains empty in each row.

There are 4 options for the first row, 3 options for the second row, 2 options for the third row and 1 option for the fourth row.

The number of options reduces to avoid more than 1 empty cell in each row or column.

Number of options is:

1/3 + 1/5 = 8/15.

step-by-step explanation:

hope it

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)