@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)
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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.
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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!

( 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

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

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.
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