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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?

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 .

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

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

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

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.

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

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

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 _________________

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.

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?

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).

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

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

_________________ IT TAKES QUITE A BIT OF EFFORT TO POST DETAILED RESPONSES. YOUR KUDOS IS VERY MUCH APPRECIATED _________________

----------------------------------------------------------------------------------------------------- IT TAKES QUITE A BIT OF TIME AND TO POST DETAILED RESPONSES. YOUR KUDOS IS VERY MUCH APPRECIATED -----------------------------------------------------------------------------------------------------

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 _________________

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. _________________

My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com

17C2 - 7C2 =136-21=115. Hand shake question always refer to a combination problems. If X men in a group then XC2 hand shakes. clearly hand shake haven't occurred between 7 so subtract 7C2.