In how many ways can you place 5 X’s and 4 O’s in a three by three board, such t
ID: 3301004 • Letter: I
Question
In how many ways can you place 5 X’s and 4 O’s in a three by three board, such that each cell gets either an X or an O? Out of these, how many of them have neither three X’s nor three O’s in any row or column?In how many ways can you place 5 X’s and 4 O’s in a three by three board, such that each cell gets either an X or an O? Out of these, how many of them have neither three X’s nor three O’s in any row or column?
In how many ways can you place 5 X’s and 4 O’s in a three by three board, such that each cell gets either an X or an O? Out of these, how many of them have neither three X’s nor three O’s in any row or column?
Explanation / Answer
There are total 9 cells available to place 5 X's and 4 O's. 5 X's can be placed in 9C5 = 126 ways (This also includes the placing of 4 O''s because remaining cells can be filled only in one way)
Lets find out the number of arrangements where 3 X's are in one row. Selection of any row can be done in 3C1 ways and now there are remaining 6 cells to places 2 X's and 3 O's. As we are considering placing only X's in a row we have to put remaining 2 X's in two different rows. This can be done in 3C1 * 2 ways (3C1 indicates selecting any cell from a row to place X)
hence possible number of ways where we have 3 X's in a row = 3C1 * 3C1 * 3C1 = 27
Now lets find out the number of arrangements where 3 O's are in one row. Note when we put 3 O's in one row there will be a row where we have 3 X's also, that is why we have excluded this condition while placing 3 X's in a row. Selecting a row for placing 3 O's in a row can be done in 3C1 ways and selecting another row to place remaining one O in 6 cells can be done in 2C1*3C1.
hence possible number of ways where we have 3 O's in a column = 3C1 * 2C1 * 3C1 = 18
Similarly in case of columns we have same number of arrangements. There are 3 arrangements which are common in rows and column where 5 X's are placed L, reverse T or _| positions.
Hence total possible arrangements where we have 3 X's or 3 O's in any row or column = (27 + 18) + (27 + 18) - 3 = 105
Required number of arrangements where neither three X’s nor three O’s in any row or column = 126 - 105 = 21