In a class of 100 students, all of them are coded with a distinct ....

Question

In a school of 900 students, all of them are coded with a distinct three-digit number from 100 to 999. If two students are select at random, what is the probability that exactly two digits of their codes are same?

A.   ^{81}C_2/^{900}C_2 
 B.   ^{90}C_2/^{900}C_2 
 C.   ^{162}C_2/^{900}C_2 
 D.   ^{243}C_2/^{900}C_2 
 E.   ^{270}C_2/^{900}C_2

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EgmatQuantExpert · Accepted Answer

Solution 
  Given:  
 • Number of students in a school = 900
• All of them are coded with a distinct number from 100 to 999
• Two students are selected at random

To find:  
 • Probability that exactly two digits of their codes are same

Approach and Working:   
 • Total number of ways of selecting 2 students from a total of 900 students =  ^{900}C_2

Now, let’s find the number of 3-digit numbers in which exactly two digits are same
 Case 1 : Units digit = tens digit
 • Units and Tens digit can take any number from 0 to 9, and the hundreds digit can take any value from 1 to 9
 o So, total such numbers = 10 * 9 = 90

• But, this includes numbers with all three digits same. So, we need to eliminate such numbers.
 o Total such numbers are {111, 222, 333, …, 999} = 9

Case 2 : Hundreds digit = tens digit
 • Hundreds and tens digit can take any number from 1 to 9, and units digit can take any value from 0 to 9
 o So, total such numbers = 10 * 9 = 90

• But, this includes numbers with all three digits same. So, we need to eliminate such numbers.
 o Total such numbers are {111, 222, 333, …, 999} = 9

Thus, total numbers in this case = 90 – 9 = 81

Case 3 : Units digit = hundreds digit
 • Units and hundreds digit can take any number from 1 to 9, and the tens digit can take any value from 0 to 9
 o So, total such numbers = 10 * 9 = 90

• But, this includes numbers with all three digits same. So, we need to eliminate such numbers.
 o Total such numbers are {111, 222, 333, …, 999} = 9

So, total such numbers = 81 + 81 + 81 = 243
 o We can select two numbers from this in  ^{243}C_2  ways

Therefore probability =  \frac{^{243}C_2}{^{900}C_2}

Hence the correct answer is Option D.

Answer: D

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chetan2u · Answer

You will have to look into the question again. How can the probability be greater than 1.
Choices C to E are all above 1. Probability >1..Impossible

Archit3110 · Answer

total students = 900

total 27 such no would be there which have two digits exactly same in each series 
for eg 100 to 200
100,101,110,112to119,121,122,131,133,141,144,151,155,161,166,171,177,181,188,191,199
so in total we have 9 such series ; 100-200,200-300,300-400,400-500,500-600,600-700,700-800,800-900,900-1000

27*9 = 243
P ()= 243c2/900c2
IMO D

EgmatQuantExpert · Answer

Hi,

There was a mistake in the question statement. We have corrected it now.

Regards,

NinetyFour · Answer

I'm guessing E. Here's my reasoning:

We have a total of 270 numbers that can have matching two digits.

100 -> 200 -> 300 -> 400....-> 900. We have a total of 10 ways we can match each number (00, 11, 22, 33... 99). Our numbers can range from 100 -> 999. This means we have 9 hundreds values we can use. We also have 3 columns that can provide the matching values (ones, tens, and hundreds).

Favorable ways we can choose two values are: 3*9*10 = 270C2

Total ways we can arrange 900 students is 900C2.

Please let me know where my logic went wrong if i made a mistake. I'm not 100% sure about this question.

sameeruce08 · Answer

i am getting 234c2/900c2

the 2 repeatative digit can be selected in 9ways and single digit must be selected apart from repeatative digit so 8ways 
thus can be arranged in3!/2!

so total arrangement is 8*9*3

no considering 0 as one digit

it can be like 101,...909 so 9 ways

when there are two zeros

number will be 100......900 again 9 ways

so total count is 8*9*3+9+9=234

234c2/900c2

help where i went wrong in counting

sameeruce08 · Answer

You are including cases where all three digits are same 
question mention two digit two be same

u sud exclude 111,222

hope that way

energetics · Answer

This question is more about number properties and combinations than probability.

Find numerator using slot method:

1) hundreds digit can be chosen in 9 ways (0 isn't permitted)
2) tens and ones digit can be chosen in 10 ways (0-9)
3) -9 for each repeat of triple digit being same (111, 222 ... 999, since we need EXACTLY 2 same) 
4) *3 for the position of the different digit
(9*10 - 9) * 3 = 81*3 = 243

fireagablast · Answer

100-900

Three slots --> 1-9, 0-9, 0-9
Total combos numbers = 9*10*10 = 900

Scenario 1: first digit and 2nd or 3rd digit match
(9*1*9) * 2! = 162 --> nine numbers for the first choice, pick the same on the 2nd, and don't pick the same on the 3rd. Multiply by 2! to get combos of 1-2 match or 1-3 match

Scenario 2: 2nd and 3rd digit match WITH ZERO
(9*1*1) = 9

Scenario 3: 2nd and 3rd digit match WITHOUT ZERO
(9*8*1) = 72 --> 2nd becomes 8 because no zero and no match with the first number

162+9+72 = 243

243 total combos of two being the same

243C2/900C2

sriharsha4444 · Answer

The question is not clear. I thought we have to pick two numbers such that exactly two digits are same between those two numbers:
i.e. 
123, 314 
in this case, 1 and 3 from first number matches 3 and 1 in second number.

But the question is about picking two numbers such that each number has exactly two digits matching within itself. This clarification may disambiguate the question

Dereno · Answer

Let us consider the three digit numbers

Hundreds  
  Tens  
  Units

The three digit numbers belong to {100 to 999}.

Thus the total number of digits = (999-100)+1 =  900 values .

We need to find : Exactly Two digits are same ?

Exactly Two digits are same = Total - (all three digits different + all three same)

Total = 900

Case 1: All three digits are different

Hundreds digit = (1 to 9) = 9 digits

Tens digit = (0 to 9) but except the hundred digit = 10-1 = 9

Unit digit = (0 to 9) but except the hundred digit and Tens digit = 10 - 2 = 8

All three digit different = 9*9*8 =  648

Case 2: All three are same

Hundred = Tens = Units

{111, 222, 333, 444, 555, 666, 777, 888, 999}

=  9 values

Exactly Two digits are same = Total - (all three digits different + all three same)

= 900 - ( 648+9)

=  243  values

so, Probability of Selecting exactly two values = (243 C 2 ) / ( 900 C 2)

^{243}C_2/^{900}C_2

Option D

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In a class of 100 students, all of them are coded with a distinct ....

Prep Toolkit

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