AnkitK wrote:

whAT is the value of (a!+b!)(c!+d!)?

1.b!d!=4(a!d!)

2.60(b!c!)=(b!d!)

@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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