Question

Only a GENIUS could solve this!!!!.....................At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?

Answer 1

here is ur answer .
there r 12 peoples...
HOPE IT HELPS YOU

Answer 2

Hi there!
Here's the answer:

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Let n members attended the party.

The first member makes (n-1) shake hands
(As he should be excluded from counting)

The second member makes (n-2) shake hands
(Excluding the first person)

:
:
:

The (n-1)th member makes [n - (n-1)] = 1 shake hands

The nth member makes n - n = 0 hand shakes

Now,

•°• Total No. of handshakes (Say N)
N= (n -1) + (n -2) + (n -3) + ……… + 1 + 0

From (n-1) we decrease by 1 & add up all terms

which is equal to :

0 + 1 + 2 + 3 + ……… + (n-1)

=> N = 1 + 2 + 3 + ……… + (n - 1)

Using sum of first 'n' Natural No.s

$Sum = \frac{n(n-1)}{2}$

Here nth term n = (n-1)

=> $N = \frac{(n-1)(n-1+1)}{2}$

=> $N = \frac{n(n-1)}{2}$

As per the given data,
N = 66

=> $\frac{n(n-1)}{2} = 66$

=> n(n-1) = 132
=> n² - n - 132 = 0

Factorise the equation

=> n² + 11n - 12n - 132 = 0
=> n(n+11) -12(n+11) = 0
=> (n+11)(n-12) = 0
=> n = -11 & n = 12

No. of Members in the party can't be -ve

•°• n = 12

Total Member in the party = 12

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Direct Formula :

No. of persons = n

Since 2 Persons are required to make a Handshake

No. of total Handshake = $n_{C}__{2}$


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Hope it helps