Question

Every body in a room shakes hands with every body else.The number of hand shake is 66 .the total number of persons in room are ? (please provide me the full described solution and please don't say that you don't know it ).

Answer 1

Answer:

12 persons

Step-by-step explanation:

let us understand this question first.

We have 5 people(say A,B,C,D,E) in a room and everyone shakes hands with each other.

now, People A shakes with 4 peoples(B,C,D,E).

People B shakes with 3 peoples(C,D,E).

People C shakes with 2 people(D,E).

People D shakes with 1 people (E).

[this is because B has already shaken hand with A, C has already shaken hand with A and B, D has already shaken hands with A,B and C and lastly,E has already shaken hand with A,B,C,D]

note:we must consider the handshakes with a particular people with other as one.that is : A shakes with B is same as B shakes with A,so we would consider this only once.

that means the total number of handshakes =4+3+2+1

=10

Generalising the above result,we can conclude that the number of handshakes= (n-1) + (n-2) + (n-3) + .....+2+1

now,this is a a.p series with first term=n-1

number of terms=n-1

common difference=-1

we know that sum of a.p series= n*(a+l)/2

where,n=number of terms

a=first term

l=last number

thus,the number of handshakes=(n-1)*(n-1+1)/2

=(n-1)*n/2

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now,let us solve the above question.

given:

(n-1)*n/2=66

or, n²-n-132=0

or, n²-(12-11)n -132=0

or, n²-12n+11n-132=0

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

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

therefore,n=12 as n≠-11

hence,the number of persons in the room is 12.