@sudhir18n: As per your request, giving the solution of this question. Your approach was correct but you overlooked one important point. Let me explain.

The question isn't as straight forward as some of you may believe. The trick here is the factorials and that n! can take only a limited set of values.

To get (a!+b!)(c!+d!), we need the values of all - a!, b!, c! and d!

Stmnt 1: b!d!=4(a!d!)
b! = 4*a!
Certainly, a can only be 3 and b can only be 4. There is no other way that this relation will hold.
Think of it this way
1*2*3*... = 4*(1*2*3*....)
4 has to be the last number multiplied on the left side.
Hence, b! = 4! and a! = 3!
No information about c and d so not sufficient.

Stmnt 2: 60(b!c!)=(b!d!)
60*c! = d!
Now this is a little trickier. c! could be 59! and d! could be 60! but
60 = 3*4*5, which leads us to another value of c and d.
c! could be 2! and d! could be 5!
3*4*5 *2! = 5!
Hence we got two different values for c! and d!
Not sufficient.

Using both statements together, we still do not know c and d, hence together also, they are not sufficient.
Answer (E)
_________________

1.
b!*d!=4*a!*d!
Dividing both sides by d! as it is +ve;
b!=4a!
Not Sufficient.

2.
60*b!*c!=b!*d!
Dividing both sides by b! as it is +ve;
60*c!=d!
Not Sufficient.

(a!+b!)(c!+d!)
=5*a!*61*c!
=305*a!*c!
Not Sufficient.

Ans: "E"

questions like these relative information among a,b,c,and d should be given.

A. b! = 4 a! provides no information regarding c! and d!. not sufficient.

B. 60c! = d! provides no information regarding a! and b!. not sufficient.

A+B

5!b * 61c! depends upon the values of b and c. hence not sufficient.

Thus E.

E . not sufficient. this is like an oral question. sub-400 kind of .

Why cant the answer be C?
Statement 1: b!=4*a!, possible only if a=3 and b=4.
Statement 2: 60*c!=d!, possible only if c=59 and d=60.
Using both statements value of given expression can be found!
Doesn't look like an easy question after all!

Why do you guys thing it cant be C

st 1. b!d!=4(a!d!) .. divide by d! we get b!=4*a!
only possibility is b =4! and a = 3!
no info on c and d .. insufficient

st2. similar solving as st.1 gives c= 59! and d =60!

together 1 and 2 we have all the information... isn't?
why is it E?

( a! + b! ) (c!+d!) = a!c!+a!d!+b!c!+b!d!

1. Not sufficient

all we know is b!=4a!

not enough information to find the given expression's value.

2. Not sufficient

all we know is c! in terms of d!

not enough information to find the given expression's value.

together

we know b! in terms of a! and
c! in terms of d!

But we still dont know anything about the values of a! and d!. so not enough

Answer is E.

Statement 1:

b!d! = 4(a!d!)
=> b! = 4 * a!
therefore, a! =3! and b! = 4!
Insufficient since we don't know c! and d!

Statement 2
60(b!c!) = (b!d!)
=> 60 c! = d!
Therefore, c!= 59! and d!= 60!
Insufficient since we don't know a! and b!

Combing 1 and 2, we get all the value and hence will be able to find the value of (a! + b!)(c! + d!)
Hence C is the answer

(a! + b!)(c! + d!)
Expanding this we get a!c!+b!d!+a!d!+b!c!............(1)

1) b!d!=4a!d!
2)60b!c!=b!d!

Treating each option individually is not gonna be fruitful in this scenario so best way is to make all the given entities into one
by combining both the options a!c!=b!d!/240

substituting the values in equation (1) :

(b!d!)/240 +b!d!+(b!d!)/4+ (b!d!)/60 this will further need the value for b!d! to solve this riddle

Hence none of the option is SUFFICIENT.

Answer:E

sorry did not get this, we already know that b!= 4! and d!= 60! so shouldn't we be able to get b!d!
please can you show how there can be values other than a!= 3!, b!= 4!,c!= 59!, d!= 60!
Thanks

Initially,i overlooked this solution but thanks for making me to dive again into this problem.
Moreover,i have referred to Krishna's (Veritas) solution while answering your question.

Taking the second part
60b!c!=b!d!
60c!=d!
now 60 = 3*4*5
two different values for c! & d!

1)c!= 59!, d!= 60!
2)c!= 2!, d!= 5!

so its doubtful to pick a specific value for factorial c & d.

Hence E

Hi! ravisinghal, I am giving it a try. Hope it will be helpful to you.


(a! + b!)(c! + d!) ?

a!c! + b!c!+ a!d! + b!d! ?

Statement 1 b!d! = 4(a!d!) or b!= 4a!

Let's put the value in question

a!c! + b!c!+ a!d! + 4(a!d!)

a!c! + b!c!+5(a!d!)
a!c! + 4a!c! + 5a!d!
5 a!c!+ 5 a!d!
5a! (c!+d!)
We don't know the value of a , c or d. Not sufficient

Statement 2 60(b!c!) = (b!d!) or 60c!= d!
a!c! + b!c!+ a!d! + b!d!
a!c! + b!c!+ a!d! + 60(b!c!)
a!c! + b!c! + 60a!c! + 60(b!c!)
61 a!c! + 61b!c!
61c! (a!+b!)

We don't know that value of c, a or b. Not sufficient

Combining statement 1 and 2,

5a! (c!+d!)
60c!= d!
5a! (c! + 60c!)
5a! * 61c!
but we still don't know the value of a and c. Not sufficient.
_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________