Eight women of different heights are at a party. Each woman decides to only shake hands with women shorter than herself. How many handshakes take place?
There are 8 women in the party, all of the different heights, and only the women-only shake hands with people shorter than them. therefore, the tallest woman does not get any handshakes. therefore, a total of 28 handshakes will take place.
7+6+5+4+3+2+1 = 28
fallest person shakes 7 hands
second to tallest person shakes 6 hands(excluding the tallest person cause otherwise it overcounts)
third to tallest person shakes 5 hands(excluding tallest and second to tallest person cause of overcounting and so on
+373 Hey there, Guest!
So, there are 8 women of different heights.
Let the eight women be A, B, C, D, E, F, G, and H.
We can assume that A is taller than B and A decides to shake hands with B, B will refuse because A is taller. If this happens across the eight women, then there will be 0 handshakes.
If this assumption is taken out, we have 7+6+5+4+3+2+1 = 28
If you still do not get it, you can look at the answer that Guest #1 copied from Brainly.
Hope this helped! :)
( ﾟдﾟ)つ Bye
+30 shortest:7
2nd shortest:6
3rd shortest:5....
tallest:0
7+6+5+4+3+2+1+0=28
I would know this literally happened to me.
+600 Here is a super easy solution:
Note that for each pair of women, one handshake takes place. The answer is then $\binom{8}{2}=28$.
Solution:
Zero (0) handshakes take place. The solution to this question is obvious and requires only logic to solve. It takes two to tango and it takes two for a handshake, and both have to be willing for that to happen.
For any combination pair of these women, one will be taller than the other and one will be shorter than the other. While the taller woman is willing to shake hands with shorter woman, the shorter woman is unwilling to shake hands with the taller woman, so she declines and no handshake occurs.
GA
