Special Coin Placement

Three Coins are randomly placed into different positions on a $4 \times 10$ grid. This means the number of all possible positions is Permutation of $40$ and $3$ or $P(40,3)$ . $P(40,3) = 40!/(40-3)! = 40!/37! = 40 \times 39 \times 38 = 59280$ . No two coins are in the same row or column means: No two coins are in the same row and No two coins are in the same column. There are $10$ rows. If you placed three coins in $10$ rows so they don't meet in the same row, there are $P(10,3)$ possible solutions or $10!/(10-3)! = 10!/7! = 10 \times 9 \times 8 = 720$ possible solutions. There are $4$ columns. If you placed three coins in $4$ columns so they don't meet in the same column, there are $P(4,3)$ possible solutions or $4!/(4-3)! = 4!/1! = 4 \times 3 \times 2 = 24$ possible solutions. That means that the number of all possible positions so no two coins are in the same row or column are $17280$ . So, the probability that no two coins are in the same row or column is $(17280)/(59280) = 72/247$ , $a/b = 72/247, a + b = 72 + 247 = 319$

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As we do with most probability problems, for this problem we'll count the number of desired outcomes and the number of total outcomes in order to come up with the probability.

Total Outcomes: There are 40 \choose 3 different ways to choose the places that the 3 coins will be. 40 \choose 3 = \frac {40 39 38}{6}

Which simplifies to just 40 13 19.

Desired Outcomes:

There are 40 possible places to put the first coin. Once the first coin is put in place, it "nullifies" all the spaces in its row and column. But how many spaces does it nullify? 10 row spaces + 4 column spaces - 1 shared row/column space = 13 nullified spaces from that first coin.

Therefore, the second coin has 40-13 = 27 places.

Similarly, once the second coin is in place then all the spaces in its row and column are nullified. That's 10 row spaces + 4 column spaces - 1 shared space = 13. However, we've also already counted some of those spaces with the first coins nullification!

In fact, exactly one column square and one row square has already been counted by the first coin, so the second coin nullifies 13 - 2 = 11 more spaces.

Therefore there are 27-11 = 16 places to put the 3rd coin.

Finally, we have 40 27 16 as the ways to put the coins, but don't forget to divide by 3! = 6 because the order of the coins doesn't matter!

so \frac {40 27 16}{6} = 40 9 8 different ways to have a desired outcome.

Now putting it all together, we have Desired Outcomes/Total Outcomes = \frac {40 9 8}{40 13 19}

= \frac {72}{247}.

72 and 247 are relatively prime, so adding them together we get 319 .

So there are a total of 4 × 10 = 40 spaces in the grid where the coins can be placed.

Now, the number of spaces where the 1st coin can be place = 40 Now the 1st coin occupies 1 row and 1 column respectively. Hence only 3 × 9 empty spaces can be ocuupied on the grid .

Therefore, the 2nd coin can be place in these 3 ×9 = 27 spaces.

Similarly the 2nd coin occupies 1 row and 1 column respectively and there are total 2 columns and 2 rows that contain any one coin. Therefore, the 3rd coin can be placed only in these 2 × 8 = 16 spaces.

Together, there are 40 × 27 × 16 cases to place these coins in such a manner. Let S=40 × 27 × 16

Now, total number of ways to place these coins in any grid are

T = 40 × 39 ×38

The probability is P = S/T

P = (40 × 27 × 16) ÷ (= 40 × 39 ×38)

Which on simplification gives P = 72 / 247 = a/b

Hence, a=72, b=247 and a+b=319

Because there are a total of 4 10=40 squares, there are a total of 40C3=9880 ways to place the 3 coins. If no 2 coins are in the same row or column, then exactly one column must be empty. (Here, we assume that there are 4 columns and 10 rows.) There are 4 ways to choose this column (There are a total of 4 columns, and you wish to choose one of them). For the first used column, there are 10 ways to choose which row it is in. For the 2nd, there are 9 (you cannot use the previous one, since all chips must be in different rows), and for the 3rd, there are 8. Therefore, our answer is (4 10 9 8)/(9880)=72/247, and the desired sum is 319.

We can place the first coin wherever we want on the board. Since we cannot put the second coin on the first coin, we have $4 \cdot 10 – 1=39$ choices to place the second coin. Of these, $4-1=3$ will be in the same column as the first coin and $10-1=9$ will be in the same row as the first coin. Thus, we have $39-3-9=27$ choices in which to put the second coin so that it will not be in the same row or column as the first coin. Thus the probability we randomly put two coins on a four by ten board that are not in the same row or column is $\frac{27}{39}=\frac{9}{13}$.

When we put the third coin down, we have $40-1-1=38$ choices. Let the first two coins be in different rows and columns. Then, there are $(10-1)+(10-1)=18$ places where the third coin would be in the same row as either the first or second coin. There are also $(4-1)+(4-1)=6$ places where the third coin would be in the same column as either the first or second coin. And of these places, 2 would place the third coin in the same row of one coin and in the same column as the other. Thus we over counted and there are only $18+6-2=22$ places where the third coin would be in the same row or column as the first two coins. Thus the probability we randomly put the third coin on a four by ten board that is not in the same row or column as two other coins (which are also not in the same row or column) is $\frac{38-22}{38}=\frac{16}{38}=\frac{8}{19}$.

Thus we need only multiply the two probabilities to get the probability the question asks for (asking yourself why? Research conditional probability). $\frac{9}{13} \cdot \frac{8}{19}=\frac{72}{247}$ Our answer is thus $72+247=319$.

Firstly we try to place 1 of the three coins.Say we choose the the upper, leftmost square.Now there are 39 available positions but the first row and column must be eliminated,thus we have * 3 9 ** squares left. Therefore,the probability of placing the second coin (say P(B) ) is 27/39 .

Similarly,the 3 rd coin can be placed in 40-2 i.e. 38 places . But the first row,first column ,second row and column has been eliminated.Thus,we have 2 8 squares left for placing the 3 * rd coin. Hence,the probability of placing the third coin (say **P(C) ) is 16/38 .

Therefore, the required probability is P(B) * P(C) . i.e. ( 27 * 16)/(39*38) which is 72/247 . Hence our answer is 72+247=319 .

When the first coin is placed, it may be placed anywhere. For the second coin to not be in the same row or column as the first coin there is a $3 \times 9$ grid where it can be placed. So the probability that it will be in a different row and column from the first is $\frac{27}{39}$ . The third coin will have a $2 \times 8$ grid where it can be placed. So the probability that it will be in a different row and column from both the first and second is $\frac{16}{38}$ . From the rule of product, we have the probability that all three coins will be in different rows and columns is $\frac{27}{39}\cdot \frac{16}{38} = \frac{72}{247}$ . So $a + b = 72 + 247 = 319$ .