How many different handshakes are possible if six girls

Question

How many different handshakes are possible if six girls are standing on a circle and each girl shakes hands with every other girl except the two girls standing next to her?

(A) 12
(B) 11
(C) 10
(D) 9
(E) 8

Had to count the scenarios, how can you calculte this with combinatorics?

Bunuel · Accepted Answer

Each girl in a circle shakes hands with 3 other girls (6 minus herself and the two girls standing next to her): 3*6=18, but since this number counts twice one handshake per pair then # handshakes possible is 18/2=9.

Answer: D.

Bunuel · Answer

Check this first: ALL YOU NEED FOR QUANT Our questions banks: viewforumtags.php From there: DS questions on Arithmetic: search.php?search_id=tag&tag_id=30 PS questions on Arithmetic: search.php?search_id=tag&tag_id=51 GMAT PS Question Directory by Topic & Difficulty: gmat-ps-question-directory-by-topic-difficulty-127957.html GMAT DS Question Directory by Topic & Difficulty: ds-question-directory-by-topic-difficulty-128728.html 100 Hardest Data Sufficiency Questions: 100-hardest-data-sufficiency-questions-162413.html 100 Hardest Problem Solving Questions: 100-hardest-problem-solving-questions-162616.html Hope this helps.

eaakbari · Answer

Well, I picked a weird way to do this but well it was a short approach

I thought of the girls as sides of a polygon, and a handshake as a diagonal.

Number of diagonal in a n-sided polygon is n(n-3)/2

Number of handshakes -> 6(3)/2 = 9

Hence D.

P.S.
Look what GMAT does to you, m picturing girls as  "sides of a polygon" !!...Sigh...

GMatAspirerCA · Answer

Hi
 I am looking for some sample questions to test my arithmatic section, Can you tell me where are these questions from? where are these sub-600, 600 to 700 questions?
thanks

GMatAspirerCA · Answer

Thanks Bunuel, your signature gives good links. thanks for that.

vietnammba · Answer

I like ur approach and this is funny lol: Look what GMAT does to you, m picturing girls as  "sides of a polygon" !!...Sigh...

Bunuel · Answer

________________ :-D

gracie · Answer

let n=6
if each of the 6 girls shook hands with each of the other 5, there would be (n)(n-1)/2=15 total handshakes
because each of the 6 shakes hands with only 3 other girls, there are only (3/5)(15)=9 total handshakes

KrishnakumarKA1 · Answer

this case is similar to find the number of diagonals = 6C1 x 3C1/2 = 9. ANSWER

lenric · Answer

Don't know whether I'm right, or wrong, but I solved this this way: 6C2 - 4C2.

Madhavi1990 · Answer

i will surely use this method for the circular/ 'standing in circle' and condition questions. Pretty insightful, thank you!

100mitra · Answer

Correct Answer - D - 9
Number of People - 6 (Condition : no hand shake with side member)

Method 1: Counting : (5+4+3+2+1) - (3+2+1) = 15 - 6= 9
Method 2 : Combinaton : 6C2 - 6C1 = 15 - 6 = 9
Method 3 : Mean :   - [(6*2)/2) = 15 - 6 = 9
Method 4 : Polygon approach : nC2-n = 6C2 - 6 = 15 - 6 = 9

sysysy9898 · Answer

i tried your method only 8 repeats came out from the 6 letters:
abcdef
a-def 
b-def 
c-def
d-abf (repeats: da db)
e-abc (repeats:ea eb ec)
f-abc (repeats:fa fb fc)

Bunuel  
Which two letters am I missing?
because the answer to me seems to be 10

ThatDudeKnows · Answer

sysysy9898

You've got incorrect letters next to each of your starting ones. No girl can shake hands with the exact same three girls as any of the others, but you have a, b, and c, each shaking hands with d, e, and f. a can't shake with f and c can't shake with d. Here are the correct groupings using your methodology.

abcdef
a-cde 
b-def 
c-efa (repeats: ca)
d-fab (repeats: da db)
e-abc (repeats:ea eb ec)
f-bcd (repeats: fb fc fd)

Paras96 · Answer

In a circle of 6 girls, each girl shakes hands with the 3 girls next to her (6 girls minus herself and the two girls on either side). This creates a total of 18 handshakes, but we need to remember that each handshake involves two individuals. So, to find the number of unique handshakes, we divide the total by 2.

Therefore, the correct number of unique handshakes among the 6 girls in the circle is 18 / 2 = 9.

Therefore, the correct number of unique handshakes among the 6 girls in the circle is 18 / 2 = 9.

nissamis00 · Answer

Look at the question closer

Its what the Number of diagonals of n sided polygon. Which in this case its n=6
so 6(6-3)/2 = 9

KarishmaB · Answer

Imagine each girl standing on the vertex of a hexagon. When one shakes hands with the girls next to her, they form a side. When she shakes hands with the girls away from her, they form a diagonal. So we need the number of diagonals that are present in a hexagon (since the girl must not shake hands with the girls next to her) Number of diagonals in a hexagon = nC2 - n Hence, Number of hand shakes here = 6C2 - 6 = 9 Check this post for a discussion on number of diagonals in a polygon: https://anaprep.com/geometry-diagonals- ... d-polygon/

orangebicycle5 · Answer

Bunuel , is this a validated solution?

Bunuel · Answer

It's not clear what 4C2 stands for in that solution, so the reasoning isn’t fully clear to me. For the correct (total) - (restriction) approach, you can refer to this explanation: https://gmatclub.com/forum/how-many-dif ... l#p3264427

How many different handshakes are possible if six girls

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How many different handshakes are possible if six girls

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