How many guests were present at the party?

(1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once
(2) If there were half as many guests present, they could sit in a row in 120 different ways

statement 1:
\(n_C_2=45\)
\(n(n-1)=90\)
n =10; sufficient

statement 2:
x! = 120; x = 5
\(\frac{n}{2}=5\); n =10
sufficient
Ans: D

(1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once
If there were 2 guests, there would be 1 handshake.
If there were 3 guests, there would be (2+1)= 3 handshakes.
If there were 4 guests, there would be (3+2+1)= 6 handshakes.
.....
If there are 10 guests, there will be (10+9+...+2+1)= 45 handshakes.
SUFFICIENT

(2) If there were half as many guests present, they could sit in a row in 120 different ways
If there were 2 guests, the guests could sit in a row in 2!=2 ways.
If there were 3 guests, the guests could sit in a row in 3!=6 ways.
.....
If there are 5 guests, the guests could sit in a row in 5!=120 ways.
THUS, if 5 is half of the number of guests, then the number of guests is 10.
SUFFICIENT

FINAL ANSWER IS (D)

Posted from my mobile device

Question: Number of guests in Party = ?

STatement (1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once

Total Number of handshakes with n people can be written in three ways







SUFFICIENT

Statement (2) If there were half as many guests present, they could sit in a row in 120 different ways
If total guests = n
then half the guests = n/2
total ways of arranging n/2 people in a row = (n/2)! = 120 (given)
but 5! = 120
i.e. n/2 = 5 i.e. n = 10
SUFFICIENT

Answer: Option D
_________________

(1) Tells us that there is a certain number of handshakes and that all the guests handshaked with everyone, so we would need to reverse use a combination of xC2=45. It is sufficient.

(2) also gives us another way of calculating the solution by telling us that (2x)!=120. It is also sufficient.

As both are sufficient, the right answer is D.

Regards,
Pablo

(1) sufic
x!/(x-2)!2!=45, x!/(x-2)!=90
x(x-1)(x-2)!/(x-2)!=90
x(x-1)=90, product of consecutive,
10(9)=90, x=10

(2) sufic
y=x/2
y!=120
3!=6, 4!=24, 5!=120
y=5, x=10

Ans (D)

(1) This is an arithmetic progression and we can use the formula ((a1+an)n)/2. ((1+x)x)/2=45 . It is obvious that we have just one unknown value. So , we can solve it. (Sufficient)
(2) This is a simple factorial (n!=120 and x=2n)and we don't need to solve it. (Sufficient)
Hence: D

Posted from my mobile device

#1

There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once

nc2=45
or say
n*(n-1)*(n-2)/2 * ( n-2) = 45
n^2-n-90=0
n=10,-9
n= 10
sufficient
#2
If there were half as many guests present, they could sit in a row in 120 different ways
n! = 120
which is when n=5
so total guest = 10
sufficeint
IMO D

How many guests were present at the party?

(1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once
(2) If there were half as many guests present, they could sit in a row in 120 different ways

How many guests were present at the party?

(1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once.

thus
\(\frac{n(n-1)}{2 }= 45\)
or n(n-1) =90
thus n =10
sufficient

(2) If there were half as many guests present, they could sit in a row in 120 different ways
(n/2)! = 120
hence n/2 = 5
or n = 10

sufficient

Each is sufficient
D
_________________

How many guests were present at the party?

(1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once
Let n be the number of guests.
Total handshakes = \(\frac{n * (n-1)}{2}\) = 45
\(n^2 - n - 90 = 0\)
n = 10
SUFFICIENT.

(2) If there were half as many guests present, they could sit in a row in 120 different ways
Let half the number of guests = m
Ways to sit in a row = m! = 120
m = 5
2m = 10
SUFFICIENT.

Answer D.
_________________

How many guests were present at the party?

(1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once
(2) If there were half as many guests present, they could sit in a row in 120 different ways.

1) Total number of guests = n. nC2 = 45. or, n!/(n-2)!2! =45 or, n(n-1)/2=45 So n is 10. Sufficient
2) (n/2)! = 120. Since 120 =5!, so there was 10 guests. Sufficient.
D is the answer.

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