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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
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16 Sep 2014, 00:35
Official Solution: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 Approach 1: Total number of handshakes possible between \(10+7=17\) people (with no restrictions) is the number 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). \(\text{Desired}=\text{Total}\text{Restriction}=C^2_{17}C^2_{7}=115\). Approach 2: Direct way: number 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: M0727
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10 Nov 2014, 09:30
For those using manhattan's notation:
10!/(2!8!) = 45
10*70 = 70
45+70=115



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18 Nov 2014, 09:40
I have a doubt on this question. It says that each business executive shakes the hand of every other B. Executive. From this statement, I understand that each B. Executvies shakes hands with 4 of the other business executives. Can someone explain me why we are considering that all B. Executives shake hands with each other?



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18 Nov 2014, 09:44
tatianamontllonch wrote: I have a doubt on this question. It says that each business executive shakes the hand of every other B. Executive. From this statement, I understand that each B. Executvies shakes hands with 4 of the other business executives. Can someone explain me why we are considering that all B. Executives shake hands with each other? Every other does not mean here every 2nd.
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18 Nov 2014, 09:48
Sorry for my ignorance, but how I am suposse to know that in this case every other, does not mean every 2nd? 'Cos in that case the answer is 90 and it is also on of the options



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04 Oct 2015, 09:12
amt88 wrote: Sorry for my ignorance, but how I am suposse to know that in this case every other, does not mean every 2nd? 'Cos in that case the answer is 90 and it is also on of the options i agree, "each other" should eliminate ambiguity
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10 Nov 2015, 21:55
shasadou wrote: amt88 wrote: Sorry for my ignorance, but how I am suposse to know that in this case every other, does not mean every 2nd? 'Cos in that case the answer is 90 and it is also on of the options i agree, "each other" should eliminate ambiguity Also got this problem wrong because of "every other" wording. Really should consider changing........
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25 Feb 2016, 13:49
I understood it the way it was intended but I can see how some people could be confused.
The way I solved was by multiplying execs by chairmen = 70
Then I thought, the first Exec can shake 9 other executives. Then the next one can shake 8, 7,6,5,4,3,2,1 which adds to 45.
70+45 = 115



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27 May 2016, 22:39
DJ1986 wrote: I understood it the way it was intended but I can see how some people could be confused.
The way I solved was by multiplying execs by chairmen = 70
Then I thought, the first Exec can shake 9 other executives. Then the next one can shake 8, 7,6,5,4,3,2,1 which adds to 45.
70+45 = 115 I used the same approach as well
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I think this is a poorquality question due to the question being worded ambiguously "each business executive shakes the hand of each of the business executives" is much clearer
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"business executive shakes the hand of every other business executive AND every chairman once"
I multipled 45 and 70 because of AND
How do we come to know whether we add or multiply ????
Pl someone help
Thanks in advance



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03 Jan 2017, 05:41
Some people may find it easier to remember and use a shortcut formula for a number of handshakes: n(n1)/2, where n is the total number of people who are going to shake hands. It is basically derived from the combinations formula n Choose 2.



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I am glad to know I am not the only one who fell into the ambiguity trap of this question. If someone is unable to write in a clear and clean way, shouldn`t be writing question at all, or at least should rely on a editor in order to avaid such gross mistake. I spent a long time trying to figure out this question because someone did a poor job writting it. I hope real GMAT Test will be more professional to this regard. And can someone please take care of correcting this mistake in order not to frustrate other people in the future? Thank you!!



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Another way to approach this is: When a group shakes hands within itself, then the number of handshakes are counted twice as when A shakes hands with B, B is also shaking hands with A. Hence, easy way is to divide the total number of handshakes by 2 within a group to avoid double counting and count total number of handshakes between different groups by simple multiplication. Here, it will be [(10*9)/2] + 10*7 = (90/2) + 70 = 115.



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Bunuel wrote: Official Solution:
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
Approach 1: Total number of handshakes possible between \(10+7=17\) people (with no restrictions) is the number 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). \(\text{Desired}=\text{Total}\text{Restriction}=C^2_{17}C^2_{7}=115\). Approach 2: Direct way: number 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 Number of handshakes for first business executive = 16 ( Business executive one shakes hand with 9 other business executives and 7 chairmen) Number of handshakes for second business executive =15 ( the handshake with the first executive is excluded here since it has been already counted for first business executive) Similarly, Number of handshakes for tenth business executive= 7 ( the handshakes with other 9 executives are excluded here since those have already been counted) Total number of handshakes = 10/2 *(16+7) = 5*23 = 115 (Sum of AP= total number of terms/2 * (first term+last term) ) Answer C
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Re: M0727
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11 Jul 2019, 11:35
Bunuel 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 Step 1: Execs shake hands everyone Ways to choose * Choices = 10 execs * 16 (they don't shake their own hand) = 160 Step 2: chairmen shake hands with execs Ways * Choices = 7 * 10 = 70 160+70 = 230, now we divide by 2 to get rid of the extras (there is no difference between person A shaking hands with B or person B shaking hands with A) 230/2 = 115.



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16 Aug 2019, 06:29
I have a lot of challenges with this sort of questions. In fact, combination and permutation type of questions. Can anyone please direct me to links on GMATCLUB that I can use to learn the fundamentals and to answer questions? Bunuel wrote: Official Solution:
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
Approach 1: Total number of handshakes possible between \(10+7=17\) people (with no restrictions) is the number 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). \(\text{Desired}=\text{Total}\text{Restriction}=C^2_{17}C^2_{7}=115\). Approach 2: Direct way: number 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 Posted from my mobile device



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Re: M0727
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16 Aug 2019, 06:37
gbengoose wrote: I have a lot of challenges with this sort of questions. In fact, combination and permutation type of questions. Can anyone please direct me to links on GMATCLUB that I can use to learn the fundamentals and to answer questions? Bunuel wrote: Official Solution:
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
Approach 1: Total number of handshakes possible between \(10+7=17\) people (with no restrictions) is the number 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). \(\text{Desired}=\text{Total}\text{Restriction}=C^2_{17}C^2_{7}=115\). Approach 2: Direct way: number 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 Posted from my mobile device21. Combinatorics/Counting Methods For more: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative MegathreadHope it helps.
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