Six girls of differing heights are arranged in 2 rows of 3 each. Each girl is taller than the girl in front of her and also taller than the girl to her right. How many arrangements of the six girls are possible?

5 possible ways

Step-by-step explanation:

Questions like this are a bit technical. Therefore, there are no perfect formula, all you have to do is picture the scene and count. Having said that. Let us use name the positions the girls are standing according to the question:

D E F

A B C

I have arranged this according to how the question has specified, meaning F is the tallest and A is the shortest if I take F to be at the back right and A to be at the front left.

In reality, A and F can never change their positions if we maintain the arrangement specified.

What we have left is to look for how to rearrange, D, E, B and C that will still make it fit the arrangement. Take note that E cannot be in front of B because she is taller than B.

Since positions for shortest and tallest are fixed, you are left with 4 people and you need two of them for the front position. This can be gotten using 4C2 = 6, but the two tall persons can't be in the front else it will disrupt the arrangement, so we have to eliminate that case which will occur once, therefore we have 5 possible ways we can make the arrangements as required.

heyyy

step-by-step explanation:

25 questions , because they start with 100.  

350-100=250 .  

. 250/10=25

step-by-step explanation:

first take off 100, then what x 10 = 250? that is your answer. (25)