Thread: fifth grade problem of the week

I am a fifth grader and am having trouble solving the problem below. Any help would be appreciated. I have worked on it and come up with an answer of 33, but my teacher says the answer is a number greater than 50 so I am stuck. Here it is:

In how many ways can seven basketball players of different heights line up in a single row so that no player is standing between two people taller than she is?

Hello, Niko!

Are you sure it's ethical for us to be helping you?
I can get you started . . .

In how many ways can seven basketball players of different heights line up in a row
so that no player is standing between two people taller than she is?

Suppose the players are: from shortest to tallest.

We have 7 spaces to fill: . _ _ _ _ _ _ _


We note that (the shortest) cannot be in the interior of the row.
. . She will be between two taller players.
Hence, she must be on the end . . . 2 choices.


We have the other 6 players to place in 6 positions:. _ _ _ _ _ _

We note that (the shortest of this group) cannot be in the interior of this row.
. . She will be between two taller players.
Hence, she must be on the end . . . 2 choices.


And so on . . . get the idea?


[I get an answer of 64.]

Thank you for your help, I wonder if it's ethical for my teacher to assign me problems that are way beyond my math level, but that you can't help me with. The advice you gave me was the way I had started the problem before e-mailing. I used the numbers 1-7 instead of letters. The shortest, #1 could be on the ends, so in two positions. The next tallest #2 could be in two positions, and then I had the following:
Player positions
3 4
4 5
5 6
6 7
7 7

So, I'm still only getting 33 ways. This is where I'm confused since you, like my teacher, say there are more.

Hello, Niko!

You and I agree on the first two players.
But I don't understand your reasoning for the others.


Call the players: . . . . from shortest to tallest.

There are seven places to fill: . _ _ _ _ _ _ _


"1" cannot be in an interior position . . . She must be on one end.
. . She has 2 choices for her end.
No matter which end she chooses, the other 6 positions are like this: ._ _ _ _ _ _

Then "2" cannot be in an interior position . . . She must be on one end.
. . She has 2 choices for her end.
No matter which end she chooses, the other 5 positions are like this:. _ _ _ _ _

Then "3" cannot be in an interior positon . . . She must be on one end.
. . She has 2 choices for her end.
No matter which end she chooses, the other 4 positions are like this: ._ _ _ _

Then "4" cannot be in an interior positon . . . She must be on one end.
. . She has 2 choices for her end.
No matter which end she chooses, the other 3 positions are like this: . _ _ _

Then "5" cannot be in the middle . . . She must be on one end.
. . She has 2 choices for her end.
No matter which end she chooses, the other 2 positions are like this: ._ _

Then "6" can choose either position: .2 choices.

Then "7" takes the remaining position: .1 choice.


Therefore, the number of choices is: .

Hi Soroban- thanks for the additional help. I now understand the problem better. I had to turn in the assignment this morning and I didn't do it correctly, but that's okay. I really wanted to know how to solve it so if I have to do something like this in the future I have a clearer idea of the steps to take.