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m07 #27

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m07 #27 [#permalink] New post 27 Jul 2009, 20:17
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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
C

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[Reveal] Spoiler: OE
Chairmen shake hands 10*7=70 with business executives times. Business executives shake hands with each other 10 C 2 times or 45 times. The total is 115 .


Can someone explain to me why its 10 C 2?
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Re: m07 #27 [#permalink] New post 11 Jul 2010, 03:14
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Another method

17 c 2 - 7 c 2 = 115

17 c 2 = total number of hand shakes between 17 people
7 c 2 = total number of hand shakes between chairmen
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Re: m07 #27 [#permalink] New post 28 Jul 2009, 06:18
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bipolarbear 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

Can someone explain to me why its 10 C 2?


10 business executives shakes hands with other 9 business executives in 10c2 ways = 45 ways
10c2 = 9+8+7+6+5+4+3+2+1 = 45

First executive shakes hands with remaining 9 executives
Second executive shakes hands with remaining 8 executives
Third executive shakes hands with remaining 7 executives
Fourth executive shakes hands with remaining 6 executives
Fifth executive shakes hands with remaining 5 executives
Sixth executive shakes hands with remaining 4 executives
Seventh executive shakes hands with remaining 3 executives
Eighth executive shakes hands with remaining 2 executives
Nineth executive shakes hands with remaining 1 executives
Tenth executive already shakes hands with all 9 executives.

7 chairmen each shake hands with all 10 executive in 10x7 - 70 ways

Total handshakes = 45+70 = 115
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Re: m07 #27 [#permalink] New post 26 Jan 2012, 01:56
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bipolarbear 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

[Reveal] Spoiler: OA
C

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[Reveal] Spoiler: OE
Chairmen shake hands 10*7=70 with business executives times. Business executives shake hands with each other 10 C 2 times or 45 times. The total is 115 .


Can someone explain to me why its 10 C 2?


Approach #1:
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.

Approach #2:
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: m07 #27 [#permalink] New post 15 Aug 2012, 04:34
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Lets think of it this way.
It takes 2 to tango, or in this case shake hands
The total number of shakehand greetings = The total Nr of 2 person groups that we can form.

10 Execs
Nr of 2 person groups we can form = 10 C 2 = 45

7 Chairman and 10 Execs
Nr of 2 person groups we can form, which INCLUDE 1 chairman and 1 Exec
7 C 1 x 10 C 1 = 70 ..

Total 70 + 45 = 115

Here is an easy way to calculate permutations and combinations without using the formula
n P M = n x n-1 x n-2 ...m times
10 P 3 = 10 x 9 x 8 ( i.e 3 terms)
11 P 6 = 11 x 10 x 9 x 8 x 7 x 6 (i.e 6 terms)


n C m = (n x n-1 x n-2 ...m times) / 1 x 2 .. m terms
10 C 2 = 10 x 9 / 1 x 2
11 C 8 = 11 C 3 = 11 x 10 x 9 / 1 x 2 x 3


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Re: m07 #27 [#permalink] New post 10 Aug 2010, 06:31
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Here's how I did it:

Scenario 1: business shakes hand only with another businessman only
#ways to select 1 businessman = 10
#ways to select another businessman = 9

#handshakes between businessmen only = 10x9/2 = 45 (We divide by 2 since order does not matter)

Scenario 2: businessman shakes hand with a chairman only
#was to select 1 businessman = 10
#ways to select 1 chairman = 7
#handshakes between businessmen and chairmen = 10x7/2 = 35

Scenario 3: chairman shakes hand with a businessman only.
#ways to select 1 chairman = 7
# was to select a businessman = 10
#handshakes between chairmen only = 7x10/2 = 35

Total handshakes = 45 + 35 + 35 = 115
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Re: m07 #27 [#permalink] New post 10 Aug 2010, 19:46
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here it is :
It takes 2 persons for one hand shake.
there are total 10 E + 7 C = 17 persons.
so total number of handshakes possible is 17C2.

The question is trying to confuse you by putting two different conditions.

17 C2 is possible when all people shakes hand with each other. But from the second condition , no chairman shakes hand with other chairman but executive.. so 7 persons are not shaking hands among themselves. So 7C2 cases must be subtracted.

Final Answer 17C2 - 7C2 = 115
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Re: m07 #27 [#permalink] New post 29 Aug 2012, 06:55
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It took me a significantly longer time than 2 mins and in the end I kind of guessed C. Honestly it was more of just a guess and not an educated guess.

Going through the explanations, it has helped me realize where I was going wrong. I was considering the total handshakes by a business executive as (9+7) and was summing this for a total of 7 executives. The explanations above really helped me understand my mistake. (Total handshakes - between chairmen) was a real smart way to go about the solution.
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Re: m07 #27 [#permalink] New post 16 Aug 2013, 01:12
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makhija1 wrote:
hello,

could someone please explain this one thing to me.

When 10 execs shake hand with each other we say 10*9/2 as the order does no matter.

But when we do 7 chairman shaking hands with 10 execs, we are saying 7*10. Why aren't we dividing this by 2?

Adi

This is essentially because executives and a chairmen are a different set of people. We are just selecting one from each group (selecting one from executives 10c1 and one from chirmen 7c1) . And hence the cases. 10c1 x 7c1. Whereas executives is a single group and we need to select two people from the same group for the handshake, hence 10c2
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Re: m07 #27 [#permalink] New post 28 Jul 2009, 06:24
Oh... thank you. I read the question as "shakes hands with every other business executive" as shaking hands with alternating executives 2, 4, 6, 8, 10... which clearly did not make sense. +1 for you
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Re: m07 #27 [#permalink] New post 11 Aug 2010, 01:54
Here is my version of explanation:

If we have 3 people (Ex: P1, P2, P3), then the possible different shakehands are:
(P1, P2)(P1,P3) (P2,P3)

If we have 4 people (Ex: P1, P2, P3, P4), then the possible different shakehands are:
(P1, P2)(P1,P3) (P1,P4) (P2, P3) (P2, P4) (P3, P4) = Choosing 2 from 4, i.e, 4C2


If each business executive shakes the hand of every other business executive ... possible ways for this scenario are: 10C2=45

and every chairman once ... possible ways for this scenario are: 10 * 7 = 70

and each chairman shakes the hand of each of the business executives but not the other chairmen ....This is already covered as part of 10C2

So, Total: 45 + 70 = 115

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Re: m07 #27 [#permalink] New post 11 Aug 2010, 07:03
I thought of it a different way...
First i broke up the people into 2 groups execs and chairmen
first the chairmen they each shake the 10 execs hands only.....10x7=70
next the execs:
they shake in sets of 2 10C2 (taking out permutations) = 45

45+70 = 115

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Re: m07 #27 [#permalink] New post 12 Aug 2010, 23:16
Its C.

10 Executives 7 Chairman

10 executives among themselves shake hands 9+8+7+6+5+4+3+2+1 = 45 times
7 chairman shakes hands with 10 executives 7*10 = 70 times

So total handshakes = 45 + 70 = 115
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Re: m07 #27 [#permalink] New post 11 Aug 2011, 04:45
Good question. C is the answer.
17C2 - 7C2 = 115
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Re: m07 #27 [#permalink] New post 11 Aug 2011, 05:58
10C2 +10x7=45+70=115
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Re: m07 #27 [#permalink] New post 03 Sep 2011, 23:13
The answer is C.

as per the above mentioned problem, here is the reasoning that will follow:
Let's assume each business exec. is assigned an alphabet A thru J
Let's assume each chairman is assigned a number 1 thru 7.

Now, starting with A (counting all busi. execs. and chairmen), A would shake 16 unique hands in all.
Next, proceeding to B (counting all busi. execs. and chairmen), B would shake 15 unique hands in all (A would be excluded).
Similarly, calculate this for C, D, E, F, G, H, I and J. Since the chairmen do not exchange handshakes among themselves, this isn't required.

Now add all the unique handshakes to find the total:
16+15+14+13+12+11+10+9+8+7 = 115
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Re: m07 #27 [#permalink] New post 04 Sep 2011, 01:57
70 handshakes from Directors to Executive
10C2 for handshakes with Exe to Exe = 45 Total Handshakes = 70+45 = 115
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Re: m07 #27 [#permalink] New post 25 Jan 2012, 22:39
bipolarbear wrote:
Oh... thank you. I read the question as "shakes hands with every other business executive" as shaking hands with alternating executives 2, 4, 6, 8, 10... which clearly did not make sense. +1 for you


That was exactly how I interpreted the question. Now I get it. :-D
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Re: m07 #27 [#permalink] New post 15 Aug 2012, 05:14
Expert's post
My take:
There are 10 executives and 7 chairman..
The given condition is that each executive can shake with another person only once.
That stands to (10*9)/2 +(10*7)/2 which equals 80.
I have divided the above by 2 to reduce redundancy, i.e., A shakes with B is the same thing as B shakes with A.
Moreover, each chair doesn't shakes with other chairman but the executives.
Hence (7*10)/2=35.
Total=80+35=115.
C
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Re: m07 #27 [#permalink] New post 15 Aug 2012, 22:34
10c2+10*7

very common kind of question
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Re: m07 #27   [#permalink] 15 Aug 2012, 22:34
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