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In a certain warehouse, each product is given a code consisting of two

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Math Expert
Joined: 02 Sep 2009
Posts: 43867
In a certain warehouse, each product is given a code consisting of two [#permalink]

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09 Mar 2017, 01:09
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65% (hard)

Question Stats:

52% (01:35) correct 48% (03:14) wrong based on 84 sessions

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In a certain warehouse, each product is given a code consisting of two letters followed by two digits. If the two letters must be identical and the product of the two digits must be even, how many possible codes exist?

A. 250
B. 650
C. 1950
D. 2600
E. 67700
[Reveal] Spoiler: OA

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Joined: 11 Aug 2013
Posts: 46
Re: In a certain warehouse, each product is given a code consisting of two [#permalink]

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09 Mar 2017, 01:18
two letters should be identical :-

first letter can be chosen in 26 ways
second in 1 way
num of ways to arrange these :- 26/2! =13

two numbers to be even : -
one has to be from 0,2,4,6,8 = 5 ways
second can be any 1 from the 10 digits = 10 ways

total ways :- 13 * 5 *10 =650...answer B
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Joined: 05 Jan 2017
Posts: 434
Location: India
In a certain warehouse, each product is given a code consisting of two [#permalink]

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09 Mar 2017, 02:56
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First two are identical letters. that means we pick onne out of 26 and place it in these two space. Number of ways = 26C1 = 26
In next two
No of two digit space filling such that ateast 1 will be even = 10*10 - 5*5 = 75
therefore total cases will be 26*75 = 1950

Option C

Last edited by ByjusGMATapp on 09 Mar 2017, 23:22, edited 1 time in total.
Intern
Joined: 21 Nov 2016
Posts: 26
Re: In a certain warehouse, each product is given a code consisting of two [#permalink]

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09 Mar 2017, 13:08
The 2 letters are identical so basically we can choose 1 (of 26 letter), which shows 26 ways. Let's see how many digit could be chosen so that their product is even.

Solution 1:
There are 10 ways to choose each digit (0-9), so the ways to choose 2 digits is 10 x 10 = 100 ways
There are 5 ways to choose odd digit (1/3/5/7/9), so the ways to choose 2 odd digits (odd product) is 5x5 = 25 ways
Therefore there are 100-25=75 ways to choose 2 digits even product

Solution 2: To double check
Even product so 2 digits could be Even Even (5*5) + Even Odd (5*5) + Odd Even (5*5). It is 3*5*5=75 ways to choose in total
Counting this way will make sure there is no redundancy/repetition.

Final answer is again 1950 (C)
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Joined: 14 May 2014
Posts: 17
Re: In a certain warehouse, each product is given a code consisting of two [#permalink]

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09 Mar 2017, 20:13
the number of ways to choose 2 identical letters = 26
the number of ways to choose 2 digits so that their product is even = 100 -25 = 75, in which:
100 is the number of ways to choose 2 any digits = 10*10
25 is the number of ways to choose 2 digits whose product is odd = 5*5 (only odd*odd results in an odd number)
Therefore, total possible codes = 26*75=1950 => Answer C
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Location: Pune, India
Re: In a certain warehouse, each product is given a code consisting of two [#permalink]

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09 Mar 2017, 21:19
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Expert's post
ravisinghal wrote:
two letters should be identical :-

first letter can be chosen in 26 ways
second in 1 way
num of ways to arrange these :- 26/2! =13

two numbers to be even : -
one has to be from 0,2,4,6,8 = 5 ways
second can be any 1 from the 10 digits = 10 ways

total ways :- 13 * 5 *10 =650...answer B

A few things to think about:

- Once you have chosen a letter in 26 ways, there is only 1 way of arranging them. e.g. if you have chosen C, there is only one way of arranging CC.
- To get an even product, it is possible that the first digit is odd and the second one is even.
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Karishma
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7951 Location: Pune, India Re: In a certain warehouse, each product is given a code consisting of two [#permalink] Show Tags 10 Mar 2017, 22:49 1 This post received KUDOS Expert's post VeritasPrepKarishma wrote: ravisinghal wrote: two letters should be identical :- first letter can be chosen in 26 ways second in 1 way num of ways to arrange these :- 26/2! =13 two numbers to be even : - one has to be from 0,2,4,6,8 = 5 ways second can be any 1 from the 10 digits = 10 ways total ways :- 13 * 5 *10 =650...answer B A few things to think about: - Once you have chosen a letter in 26 ways, there is only 1 way of arranging them. e.g. if you have chosen C, there is only one way of arranging CC. - To get an even product, it is possible that the first digit is odd and the second one is even. Responding to a pm: Then how do you please explain questions like arranging Mississippi etc..where identical objects to be divided... similar case with CC also..?? Ways of arranging ABC = 3! ABC ACB BAC BCA CAB CBA Ways of arranging AABC = 4!/2! Assume the two As are different - $$A_1$$ and $$A_2$$. Ways of arranging = 4! But the two As are the same. So $$A_1BCA_2$$ is the same as $$A_2BCA_1$$ etc. So divide by 2. By the same logic: Ways of arranging CC = 2!/2! = 1 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: In a certain warehouse, each product is given a code consisting of two   [#permalink] 10 Mar 2017, 22:49
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