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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!

garimavyas wrote:

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

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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WE: General Management (Non-Profit and Government)

Re: What is the value of (a! + b!)(c! + d!)? [#permalink]

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10 Nov 2014, 05:49

(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

Re: What is the value of (a! + b!)(c! + d!)? [#permalink]

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12 Jan 2015, 15:20

akumar5 wrote:

(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

WE: General Management (Non-Profit and Government)

Re: What is the value of (a! + b!)(c! + d!)? [#permalink]

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12 Jan 2015, 21:50

1

This post was BOOKMARKED

qlx wrote:

akumar5 wrote:

(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.

Re: What is the value of (a! + b!)(c! + d!)? [#permalink]

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17 Oct 2017, 08:33

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