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

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In a certain warehouse, each product is given a code consisting of two [#permalink]

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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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Re: In a certain warehouse, each product is given a code consisting of two [#permalink]

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New post 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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In a certain warehouse, each product is given a code consisting of two [#permalink]

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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.
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Re: In a certain warehouse, each product is given a code consisting of two [#permalink]

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New post 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

Final answer: 75*26=1950 (C)

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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Re: In a certain warehouse, each product is given a code consisting of two [#permalink]

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New post 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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Re: In a certain warehouse, each product is given a code consisting of two [#permalink]

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New post 09 Mar 2017, 21:19
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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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Re: In a certain warehouse, each product is given a code consisting of two [#permalink]

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New post 10 Mar 2017, 22:49
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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
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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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