Asked by anu
there are 12 people at a party. each person shakes hands with each of th other guests. how many hadshakes will there be in all
Answers
Answered by
Reiny
each of the 12 shakes hands with 11 others, but A shaking with B is the same as B shaking with A, so
12 x 11/2 = 66
In combination notation:
C(12,2) = 12!/(10!2!) = 66
12 x 11/2 = 66
In combination notation:
C(12,2) = 12!/(10!2!) = 66
Answered by
tchrwill
Another viewpoint:
A shakes the hand of 11 people.
B shakes the hand of 10 people, already having shaken the hand of A.
C shakes the hand of 9 people, already having shaken the hand of A and B.
Therefore, the total number of hand shakes is simply the sum of the numbers from 1 through 11 or, S = n(n + 1)/2 = 11(12)/2 = 66.
A shakes the hand of 11 people.
B shakes the hand of 10 people, already having shaken the hand of A.
C shakes the hand of 9 people, already having shaken the hand of A and B.
Therefore, the total number of hand shakes is simply the sum of the numbers from 1 through 11 or, S = n(n + 1)/2 = 11(12)/2 = 66.
Answered by
Anonymous
1,993,373
Answered by
Jon
You might want to recheck the math. If you start with #12 that's 11 shakes then 10, 9,8,7,6,5,4,3,2. when you get to #2 he shakes with #1 that is 65 #1 has shook with everybody so doesn't need to shake with anyone.
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