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 January 21st, 2015, 05:02 PM #1 Newbie   Joined: Jan 2015 From: Victoria Posts: 1 Thanks: 0 Permutation and Combinations Problem I am trying to do an assignment for my current math class, and I have been stuck on it for hours. Here is the question: the 6 digits 0 through 5 can be arranged in a 2X3 table (a table with 2 rows and 3 columns). How many ways are there to do this if for each column, the entry in row 1 must be less than the entry in row 2? the answer is 90, but I can't for the life of me figure out how. January 21st, 2015, 05:35 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra Let's start by picking three numbers for the bottom row: $\{b_1, b_2, 5\}$ in ascending order. We'll then arrange the remaining three numbers above them so that we have three columns. Finally, we'll count the possible arrangements of the columns. If $b_2 \ne 4$ we must have $b_2 = 3$ (the four must go above the 5 and then there is no number that can go below the 3). So there are only two options for $b_2$. If $b_2 = 4$ then $b_1$ can be any of 1, 2 or 3.If $b_1 = 3$ we can assign 0, 1 and 2 in any order to $a_1$, $a_2$ and $a_3$. There are 6 ways to order them. If $b_1 = 2$, the 3 must go above either the 4 or the 5 (two choices) and the 0 and the 1 can be placed anywhere in the other two places (two choices). That's 4 ways to do it. If $b_1 = 1$, the 0 must go above the 1 and then the other two can go in either of the remaining places. That's 2 ways.That gives a total of 12 ways to have 4 and 5 on the bottom. If $b_2 = 3$ then $b_1$ can be either 1 or 2. That's two choices.If $b_1 = 2$ we must have the 4 above the 5 and the 0 and 1 can then be placed anywhere. So that is 2 ways to do it. If $b_1 = 1$, the 0 must go above it and the 4 must go above the 5 so the 2 goes above the 3. That's 1 more way to do it.That's 3 ways to have 5 but not 4 on the bottom. Now we have 12+3=15 ways to create three columns. All we have to do is to arrange the columns. There are 6 ways to arrange the columns, so that's 6*15 = 90 ways to fill the grid as required. Of those six ways, Thanks from jdesireg January 21st, 2015, 07:26 PM #3 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond There are 5 choices for the top row: {0, 1, 2}, {0, 1, 3}, {0, 1, 4}, {0, 2, 3} and {0, 2, 4}. Note that these rows can be arranged in 3 nPr 3 = 6 ways. For the table using the first top row, the numbers in the rows can be arranged in any order so we have 6 * 6 = 36 possible arrangements. For the table using the second top row, there are two places to place the 2 in the bottom row and two places to place the 4 and the 5 (for each positioning of the 2), so we have 6 * 2 * 2 = 24 possible arrangements. For the table using the third top row, there are two ways to arrange the bottom row ({2, 3, 5} or {3, 2, 5}, for example), so we have 6 * 2 = 12 possible arrangements. For the table using the fourth top row, there are two ways to arrange the bottom row so we have 6 * 2 = 12 possible arrangements. Finally, for the fifth top row, the bottom row can be arranged in only one way, so we have 6 possible arrangements. 36 + 24 + 12 + 12 + 6 = 90. January 21st, 2015, 07:27 PM   #4
Math Team

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Hello, jdesireg!

Quote:
 The six digits 0 through 5 are arranged in a 2x3 table. How many ways are there to do this if for each column, the entry in row 1 must be less than the entry in row 2? The answer is 90.

Suppose the first column is ${0\choose1}:\;\;\boxed{\begin{array}{ccc}0&*&* \\ 1&*&* \end{array}}$

The other four digits {2, 3, 4, 5} can be placed in three ways:

$\quad \boxed{\begin{array}{ccc}0&2&4 \\ 1&3&5 \end{array}}\qquad \boxed{\begin{array}{ccc}0&2&3 \\ 1&4&5 \end{array}} \qquad \boxed{\begin{array}{ccc}0&2&3\\ 1&5&4\end{array}}$

There are five choices for the first column: $\:{0\choose1},\;{0\choose2},\;{0\choose3},\;{0 \choose4},\;{0\choose5}$

The other two columns can be filled in three ways.

Then the three columns can be permutated in $3!=6$ ways.

Answer: $\:5\cdot3\cdot6 \:=\:90$ ways. January 22nd, 2015, 03:04 AM #5 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 The numbers in each of the columns can be arranged in 2 ways. So in three columns in 2^3 = 8 ways. The 6 numbers can be placed in 6! = 720 ways, of which 1/8 of them are as desired. So (720/8 = 90) ways. Thanks from greg1313 Tags combinations, permutation, problem Search tags for this page

### grid problems in permutations and combinations

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