Konstantin Lynov wrote:
Eight people are to make photo, four of them being standing in the first row and four of them in the second. Every man standing in the front row should be lower that that standing exactly behind him, and all people from the same row should be arranged in order of height increase from left to right. Providing the restrictions mentioned above how many different photos could be taken?
I don't know if there is a formula, so I will try to solve this logically:
First, let's say the formation looks like this:
where the heights go from 1 to 8 and the unknowns are 0's.
We know that at a minimum, it must look like this:
Now, consider the person top row 2nd from left. He must have one person shorter than him both to his left and below. So at a minimum, he must be a 4. Similarly, the person in the top row, 3rd from the left must have 5 shorter than him (two on his left, one below plus the two to the left of the one below) so his minimum height is 6.
Now we can count. Let's designate the arrangement as:
Y can only be 4,5, or 6.
If Y is 4, Z can be 6 or 7 and X can be 2 or 3. 4 ways
If Y is 5, Z can be 6 or 7 and X can be 2, 3, or 4. 6 ways.
If Y is 6, Z must be 7 and X can be 2, 3, 4, or 5. 4 ways.
For each of the combinations above, the bottom row is determined (we have 3 numbers left all of which must be in order).
Hence, there are 4 + 6 + 4 = 14 ways.
KL, is there a better way to solve this?
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993