Can you shakehands with your friends

Everybody in a room shakes hands with everybody else. The total number of handshakes is 66. Find the number of people in the room.


The answer is 12.

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9 solutions

If there are 'n' people.

No. of handshakes = (n*(n-1))/2

Hence, n=12.

Prasun Biswas
Mar 23, 2014

This is actually a combinatorics problem. I don't know what the HELL is it doing here in Number Theory. Anways, let us take the no. of people as n n . Now, we know that in a handshake, there are 2 2 people and the arrangement of the people in a handshake doesn't matter, i.e., if the 2 2 person's position are exchanged, then also the handshake is the same as the previous one and handshakes are not double counted like that. So, here we use combinations to find out the no. of people.

From all the above data, we can say that there are n C 2 nC2 or ( n 2 ) \binom{n}{2} handshakes as for a handshake, we select 2 2 people out of n n people. Also, it is said that there had been 66 66 handshakes. So --->

( n 2 ) = 66 \binom{n}{2}=66

n ! 2 ! × ( n 2 ) ! = 66 \implies \frac{n!}{2!\times (n-2)!}=66

n ! ( n 2 ) ! = 66 × 2 ! = 66 × 2 = 132 \implies \frac{n!}{(n-2)!}=66\times 2! = 66\times 2 = 132

n ( n 1 ) = 132 \implies n(n-1)=132

n 2 n 132 = 0 \implies n^2-n-132=0

( n 12 ) ( n + 11 ) = 0 \implies (n-12)(n+11)=0

n = 12 o r n = ( 11 ) \implies n=12\quad or \quad n=(-11)

But, no. of people cannot be (-ve), so n ( 11 ) n\neq (-11) , so we finally have n = 12 n=\boxed{12}

So, there are 12 persons in the room.

P.S. -- For those who dont know about combinations, I am giving a little tutorial here. Combinations is referred to the no. of ways in which r r things can be selected out of n n things at a time and that value is represented by n C r nCr or ( n r ) \binom{n}{r} .

We have the formula as follows ---> n C r = ( n r ) = n ! r ! × ( n r ) ! nCr=\binom{n}{r}=\frac{n!}{r!\times (n-r)!}

What's (-ve)?

. . - 4 months ago
Rahma Anggraeni
May 19, 2014

n ! 2 ! ( n 2 ) ! = 66 \frac{n!}{2!(n-2)!}=66

n ! ( n 2 ) ! = 132 \frac{n!}{(n-2)!}=132

n ( n 1 ) ( n 2 ) ! ( n 2 ) ! = 132 \frac{n(n-1)(n-2)!}{(n-2)!}=132

n ( n 1 ) = 132 n(n-1)=132

n 2 + n 132 = 0 n^{2}+n-132=0

( n 12 ) ( n + 11 ) = 0 (n-12)(n+11)=0

n = 12 n=12 or n = 11 n=-11 , we choose the real number 12 \boxed{12}

S W
Apr 21, 2014

(n(n-1))/2 =66 , hence n=12

Diyanko Bhowmik
Apr 4, 2014

No. of handshakes = n ( n 1 ) 2 = 66 \frac { n(n-1) }{ 2 }=66 . . . So we get the quadratic equation of n 2 n 132 { n }^{ 2 }-n-132 . . .which gives n = 12 and -11 . . . So, the answer is 12.

Lira Zabin
Mar 21, 2014

12C2=66

ufff!! ami bujhina tmi eto simply kvabe koro??! ami to onk ghuray pechay answr kri... :/ tmi aslei chorom!! (y)

Md. Mohaiminul Islam - 7 years, 2 months ago
Sathi Nagi Reddy
Mar 13, 2014

its just c(n,2) n*(n-1)/2!=66 n=12

let no. people be x then (x*x-1)/2 =12

Ayush Anand
Mar 1, 2014

let the no. of people be x then , No. of handshakes= (x*(x-1))/2) therefore x equals 12

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