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

Question

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

Show Answer

Official Answer

C

A. 250
B. 650
C. 1950
D. 2600
E. 67700

phuongnina · Accepted Answer

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)

ravisinghal · Answer

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

KrishnakumarKA1 · Answer

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

trangphamthuy91 · Answer

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

KarishmaB · Answer

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.

KarishmaB · Answer

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

KarishmaB · Answer

Responding to a pm:

For the product to be even, at least one digit has to be even. It can be that first digit is even and second odd or the second digit is even and first odd or both are even.

So number of ways to get even product =
5 * 5 (first digit even and second digit odd) + 5*5 (first digit odd and second digit even) + 5*5 (both digits even) = 75

Possible number of distinct codes = 26 * 75 = 1950

siddhantvarma · Answer

Hey  KarishmaB 
This is what I did to solve this and I want to understand where I'm going wrong:

First two letters are identical = 26C1
Arrangements don't matter here since both are identical

Now for the two digits we have even*odd + odd*even + even*even
An even integer can be selected in 5C1 ways
Similarly, an odd integer can be selected in 5C1 ways.

For even*odd = 5C1*5C1 = 25
For odd*even = 5C1*5C1 = 25
I can either take both the above cases, or take just even*odd and multiple it by 2! to get the arrangements. Regardless the result here will be 25+25 = 50
Hence 50 ways to select and arrange one even and one odd

Now for even*even = 5C1*5C1 = 25
However, to account for arrangements I'll multiply the above by 2!
For instance, if I select 2 and 4, there will be an arrangement with 4 and 2 as well so each selection will have 2! arrangements
Hence 25*2 = 50 ways to select and arrange both even

This way the result becomes: 26*(50+50) = 2600

Where am I going wrong in my approach? What am I overcounting?

KarishmaB · Answer

This is not correct. You will not multiply by 2. 
5C1*5C1 gives you the case of 24 as well as 42. No arrangement required.

eshareddy2016 · Answer

Hi All, Can someone please explain how can zero be chosen in any of the combinations given that the product when zero is chosen is zero & zero is neither even nor odd- and the question explicitly states the product should be even. So afaik, 0 should be left out of all selections.

Bunuel · Answer

Zero is an even integer.

ZERO:

1. Zero is an  INTEGER .

2. Zero is an  EVEN  integer.
 
 3. Zero is  neither positive nor negative  (the only one of this kind)

4. Zero is  divisible by EVERY integer except 0 itself  ( \frac{x}{0} = 0 , so 0 is a divisible by every number, x).

5. Zero is a  multiple of EVERY integer  ( x*0 = 0 , so 0 is a multiple of any number, x)

6. Zero is  NOT a prime number  (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed:  anything/0 is undefined .

8. Any non-zero number  to the power of 0 equals 1  ( x^0 = 1 )

9.  0^0  case is  NOT tested on the GMAT .

10. If the exponent n is positive (n > 0),  0^n = 0 .

11. If the exponent n is negative (n < 0),  0^n  is undefined, because  0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0} , which is undefined.  You CANNOT take 0 to the negative power .

12.  0! = 1! = 1 .

SanuBest · Answer

KarishmaB

But how does it account for 92 ? Please help me understand, Im confused

NuttyIan · Answer

I got this by tackling the last part first, since every letter needs to be the same and there are 26 letters in the latin alphabet, it is 26 choices. We multiply that by how many even products are there. I started with 1s in the 10s column but then I realized there are also zeros to start. So the zeros are all an option since any number times zero is an even zero. Looping back to the 1s we can see that even x odd is even, while odd x odd is still odd. Since half the numbers in the 1s column are odd, there are 5 even numbers. Then I worked my way to the 2s in the 10s column, where the pattern clicked in my brain. All 2's numbers are even. So for every 20 numbers, 15 are even. 100 total /20 = 5 times this pattern occurs. 5*15 =75, Bringing back in 26 so 26* 75, at this point you don't need to multiply the numbers since the only number that seems right is C, 1900.

Easier visualization:
0 times x = all zero (10 numbers)
1 times x= all even x numbers are even (5 numbers)
2 times x= all even (10 numbers)
3 times x= all even x numbers are even (5 numbers)...
and so on until we stop at 99.

Easier visualization:
0 times x = all zero (10 numbers)
1 times x= all even x numbers are even (5 numbers)
2 times x= all even (10 numbers)
3 times x= all even x numbers are even (5 numbers)...
and so on until we stop at 99.

KarishmaB · Answer

There are 10*10 = 100 combinations of 2 digits possible.
To get an even product, you need at least one digit to be even. To get an odd product, you need both digits to be odd. So you can make an odd product in 5*5 = 25 ways. This includes all combinations such as 17, 71, 39, 93 etc. since both places can take any one of the 5 digits (1/3/5/7/9).
All other 75 combinations will have an even product because at least one of their digits will be even.

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

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