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29 Jul 2022, 01:30
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The numbers 1, 2, ..., 9 are randomly placed into the 9 squares of a 3*3 grid. Each square gets one number, and each of the numbers is used once. What is the probability that the sum of the numbers in each row and each column is odd?
(A) \(\frac{1}{21}\)
(B) \(\frac{1}{14}\)
(C) \(\frac{5}{63}\)
(D) \(\frac{2}{21}\)
(E) \(\frac{1}{7}\)
Are You Up For the Challenge: 700 Level Questions
(A) \(\frac{1}{21}\)
(B) \(\frac{1}{14}\)
(C) \(\frac{5}{63}\)
(D) \(\frac{2}{21}\)
(E) \(\frac{1}{7}\)
Are You Up For the Challenge: 700 Level Questions
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29 Jul 2022, 08:18
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Bunuel
The numbers 1, 2, ..., 9 are randomly placed into the 9 squares of a 3*3 grid. Each square gets one number, and each of the numbers is used once. What is the probability that the sum of the numbers in each row and each column is odd?
(A) \(\frac{1}{21}\)
(B) \(\frac{1}{14}\)
(C) \(\frac{5}{63}\)
(D) \(\frac{2}{21}\)
(E) \(\frac{1}{7}\)
(A) \(\frac{1}{21}\)
(B) \(\frac{1}{14}\)
(C) \(\frac{5}{63}\)
(D) \(\frac{2}{21}\)
(E) \(\frac{1}{7}\)
Let's look at the Odds.
How many total ways are there to place the Os?
There are five of them and there are nine spaces into which we could place them.
So, there are 9C5 = \(\frac{9*8*7*6}{4*3*2} = 126\)
How many ways are there to win?
Each column must be either OOO or OEE.
We have five Os, so once we place one into each column, we still have two to place. Those two must both go in the same column.
The same is true of the rows.
There are 3 ways to select which column and 3 ways to select which row.
\(3*3 = 9\)
\(\frac{9}{126} = \frac{1}{14}\)
Answer choice B.
29 Jul 2022, 11:19
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If you map out the odds and evens on the grid, you will see that a proper solution requires a column of 3 odds and a row of 3 odds.
There are 3*3=9 ways this can be done.
The odds can be permuted 5! ways and the evens 4! ways for a total of:
9*5!*4!.
The total number of ways the 9 cells can be permuted is
9!
So the probability is
9*5!*4!/9! =
9*5*4*3*2*4*3*2/9*8*7*6*5*4*3*2 which equals
4*3*2/8*7*6 = 3/42 = 1/14
Posted from my mobile device
There are 3*3=9 ways this can be done.
The odds can be permuted 5! ways and the evens 4! ways for a total of:
9*5!*4!.
The total number of ways the 9 cells can be permuted is
9!
So the probability is
9*5!*4!/9! =
9*5*4*3*2*4*3*2/9*8*7*6*5*4*3*2 which equals
4*3*2/8*7*6 = 3/42 = 1/14
Posted from my mobile device
