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10 business executives and 7 chairmen meet at a conference

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10 business executives and 7 chairmen meet at a conference [#permalink]

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10 business executives and 7 chairmen meet at a conference. If each business executive shakes the hand of every other business executive and every chairman once, and each chairman shakes the hand of each of the business executives but not the other chairmen, how many handshakes would take place?

A. 144
B. 131
C. 115
D. 90
E. 45
[Reveal] Spoiler: OA

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Re: 10 business executives and 7 chairmen [#permalink]

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manalq8 wrote:
10 business executives and 7 chairmen meet at a conference. If each business executive shakes the hand of every other business executive and every chairman once, and each chairman shakes the hand of each of the business executives but not the other chairmen, how many handshakes would take place?


144
131
115
90
45
what is the quickest way to solve it. help please


Total # of handshakes possible between 10+7=17 people (with no restrictions) is # of different groups of two we can pick from these 10+7=17 people (one handshake per pair), so \(C^2_{17}\). The same way: # of handshakes between chairmen \(C^2_{7}\) (restriction).

\(Desired=Total-restriction=C^2_{17}-C^2_{7}=115\).

Or direct way: # of handshakes between executives \(C^2_{10}\) plus 10*7 (as each executive shakes the hand of each 7 chairmen): \(C^2_{10}+10*7=115\).

Answer: C.
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Re: 10 business executives and 7 chairmen meet at a conference [#permalink]

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New post 17 Jan 2012, 03:45
My answer is also 115. solved it in generic way. Bunnel thanks for your explanation.

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New post 19 Jan 2012, 00:06
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Business Executives (BE) can shake hands with every other business executives and chairman (C).

BE and C - 10 *7 =70
Be vs BE - 10c2 = 10!/(2!*(10-2)!)= 45
So total hand shakes = 70 +45 =115
Since chairman shakes hands only with business executive, it is enough if we compute that once

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Re: 10 business executives and 7 chairmen meet at a conference [#permalink]

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New post 19 Jan 2012, 10:55
thank everyone, appreciate it
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Re: 10 business executives and 7 chairmen meet at a conference [#permalink]

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New post 24 Dec 2012, 01:00
For each business executive, he will shake a total of

\(9 + 7 = 16 shakes\)

Total Business shakes ->\(16 *10 / 2 = 80\)

For each chairman, he will shake a total of

\(10 shakes\)

Total chairman shakes -> \(10 *7 / 2 = 35\)

Hence total shakes -> \(80 + 35 = 115\)
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Re: 10 business executives and 7 chairmen meet at a conference [#permalink]

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New post 28 Dec 2012, 01:25
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manalq8 wrote:
10 business executives and 7 chairmen meet at a conference. If each business executive shakes the hand of every other business executive and every chairman once, and each chairman shakes the hand of each of the business executives but not the other chairmen, how many handshakes would take place?

A. 144
B. 131
C. 115
D. 90
E. 45


How many times to select 2 persons to handshake from 10 execs? \(\frac{10!}{2!8!} = 45\)
How many handshakes by 10 exec to all chairment each? \(10*7 = 70\)
How many times each chair to handshake 10 execs? This has already been counted above.

\(45 + 70 = 115\)

Answer: C
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Re: 10 business executives and 7 chairmen meet at a conference [#permalink]

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New post 24 Nov 2013, 04:41
Well this is a long shot method, but if you are like me and have problem calculating permutations and combination this may help:

Now the question requires us to calculate the total number of handshakes. The total number of hand shakes is equal
(Unique)Hand Shakes made by the Executives + (Unique) Hand Shakes Made by the CEOs

So to Calculate the Unique Handshakes made by the executives, the following calculations need to be made:

1st Executive shakes hand with Executives (A)= 9, He shakes hands with CEOs (B)= 7, Total Hand Shakes=A+B = 16
2nd Executive shakes hand with Executives (A)= 8, He shakes hands with CEOs (B)= 7, Total Hand Shakes=A+B = 15
(The number has decreased because his hand shake with the 1st executive cannot be recounted)
3rd Executive shakes hand with Executives (A)= 7, He shakes hands with CEOs (B)= 7, Total Hand Shakes=A+B = 14
4th Executive shakes hand with Executives (A)= 6, He shakes hands with CEOs (B)= 7, Total Hand Shakes=A+B = 13
5th Executive shakes hand with Executives (A)= 5, He shakes hands with CEOs (B)= 7, Total Hand Shakes=A+B = 12
6th Executive shakes hand with Executives (A)= 4, He shakes hands with CEOs (B)= 7, Total Hand Shakes=A+B = 11
7th Executive shakes hand with Executives (A)= 3, He shakes hands with CEOs (B)= 7, Total Hand Shakes=A+B = 10
8th Executive shakes hand with Executives (A)= 2, He shakes hands with CEOs (B)= 7, Total Hand Shakes=A+B = 09
9th Executive shakes hand with Executives (A)= 1, He shakes hands with CEOs (B)= 7, Total Hand Shakes=A+B = 08
10th Executive shakes hand with Executives (A)= 0, He shakes hands with CEOs (B)= 7, Total Hand Shakes=A+B = 07

Now, the Unique Handshakes made by the CEOs is 0. Because all their handshakes are with the Executives, that have been covered earlier. So the total handshakes in the given situation are equal to the sum of the unique handshakes made by the executives which are 115. And our answer is C.

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Re: 10 business executives and 7 chairmen meet at a conference [#permalink]

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New post 12 Dec 2013, 23:48
1)10C2=45 (ways of picking 2 people out of 10 business executives, which equals to number of their handshakes)
2)10*7=70 (every business executive has 7 chairmen to shake hands)
3)70+45=115

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Re: 10 business executives and 7 chairmen meet at a conference   [#permalink] 27 Jan 2017, 03:44
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