The numbers 1, 2, 3, 4, 5, and 6 are to be entered into the boxes on top, so that there is one number per box, and each number is used exactly once. In addition, in each row, the numbers increase from left to right, and in each column, the numbers increase from top to bottom. Find the number of possible placements. By the way, the answer is not 54 or 162
Based on my past answer of 54 (you posted the same question earlier and I incorrectly answered 54), I noticed that there are only 2 cases that don't fit.
So 54 - 2 = 52
So 52 is the answer?
How many ways can 3 numbers be chosen from 6? 6C3=20 ways
So there are 20 ways to chose two sets of 3 numbers.
Put each triplet in increasing order and put the one with she smallest number in the top left corner.
Now only half of those will have the second column in ascending order so that cuts it down to 10ways
But only half of those 10 will have the third culumn in ascending order so
There will be 5 possible outcomes.
Melody you are correct that the number of ways to choose 3 numbers out of 6 is 20 but the number of ways to place 1-6 in the boxes such that they are ordered from left to right AND such that the smallest number (1) is in the top left is not 20, it is 10.
My final answer was 5, not 20.
I just worked out the answer manually.
My logic was a little off but the answer was still 5
I know 5 was your final answer, I just pointed out that to get that you divided 20 by 2 twice because you assumed that the number of ways to place 1-6 in the boxes such that they are ordered from left to right and such that the least number would be in the top left corner (ignoring columns being ordered from top to bottom) is 20, divided by 2 to find the number of such placements where the 2nd column is ordered and again to find the number of placements where both 2nd and third are ordered.