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In a class comprising boys and girls, there were 45 hand

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In a class comprising boys and girls, there were 45 hand [#permalink] New post 02 Mar 2006, 03:17
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In a class comprising boys and girls, there were 45 hand shakes amongst the girls and 105 hand shakes amongst the boys. How many hand shakes took place between a boy and a girl, if each member of the class shook hands exactly once with every other student in the class?

25
15
150
300
24
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 [#permalink] New post 02 Mar 2006, 07:50
I got 150.

I did trial and error to get 45 handshakes among X girls. 45=XC2. After a couple of tries, I got X=10.

For Y boys, 105=YC2. So, Y=15.

Between boys and girls is X*Y=10*15=150.
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 [#permalink] New post 02 Mar 2006, 08:51
Logically speaking this is what I get.

No of Girls = n(n+1)/2 = 45, hence n = 9
No of Boys = n(n+1)/2 = 105. hence n = 14

Handshakes between a boy and a girl = 14*9 = 126.
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 [#permalink] New post 02 Mar 2006, 09:05
g = # of girls; b=#of boys
gC2 = 45
therefore...
g!/2!(g-2)! = 45
g(g-1)!/2! = 45
g(g-1) = 45 x 2
g^2 - g = 90.......resolves to 10

after following same steps..
b^2 - b = 210......resolves to 15

handshakes btw boys and girls = 15 x 10 = 150
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 [#permalink] New post 02 Mar 2006, 09:12
Gordon wrote:
Logically speaking this is what I get.

No of Girls = n(n+1)/2 = 45, hence n = 9
No of Boys = n(n+1)/2 = 105. hence n = 14

Handshakes between a boy and a girl = 14*9 = 126.

You method is a much quicker way to solve the problem but your mistake is:
# of Girls should be n(n-1)/2 = 45 and # of boys should be n(n-1)/2 = 105

Last edited by trublu on 02 Mar 2006, 10:21, edited 1 time in total.
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 [#permalink] New post 02 Mar 2006, 09:32
trublu my friend, u got me out of potential trouble:). Thanks for pointing that out.
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 [#permalink] New post 03 Mar 2006, 12:58
Hey Guys...this is really one of my weaknesses. Could you explain the logic for setting up the equation n(n-1)/2.... :lol:
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C if this helps [#permalink] New post 04 Mar 2006, 00:31
If there are 10 girls, the handshakes go as :
1st girl = 9
2nd girl = 8
.....
10th girl = 0
So, it is 9+8+.....+1+0 = Sum of natural numbers upto 9 = 9*10/2 = 45

Now, if we use the formula of sum of natural numbers taking N=10,
then, N (N+1)/2 , for 10 girls becomes 10 * 11 / 2 = 55.

Whereas, the question asked is about handshakes between the girls and not the girls,
we need to consider 9 + 8 +....... + 1
which boils down to N (=9 ) * N+1 (=10) / 2

But, if we want a direct answer, we can take N as 10 girls, in which case
formula has to be considered as N(N-1)/2.
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 [#permalink] New post 04 Mar 2006, 10:19
positive soul wrote:
Hey Guys...this is really one of my weaknesses. Could you explain the logic for setting up the equation n(n-1)/2.... :lol:


that is pretty simple just think about adding the numbers from 1 to 100

then you can add 1+100 = 101
2+99 = 101
3+98 = 101
.....
50+51 =101

so you have 50 pairs that add 101

so for the number 100

n=100

then n(n+1)/2 = 50*(101)

some people say that Gauss found out this when he was 5 years old
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 [#permalink] New post 05 Mar 2006, 04:33
positive soul wrote:
Hey Guys...this is really one of my weaknesses. Could you explain the logic for setting up the equation n(n-1)/2.... :lol:


We can get this equation by normal combination ...

Handshakes among girls = 45
Let number of girls be = n
=> so, total handshakes are nC2 = n!/[(n-2)! * 2!] = n*(n-1)/2

i.e. n*(n-1)/2 = 45. then solve for n & we get n = 10

Similarly for boys, m*(m-1) = 105. then solve for m & we get m = 15

So answer is, 15*10 = 150
  [#permalink] 05 Mar 2006, 04:33
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