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In a class of 100 students, all of them are coded with a distinct ....
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Updated on: 30 Jan 2019, 10:46
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Re: In a class of 100 students, all of them are coded with a distinct ....
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30 Jan 2019, 04:01
EgmatQuantExpert wrote: In a class of 100 students, all of them are coded with a distinct threedigit number from 100 to 199. If two students are select at random, what is the probability that exactly two digits of their codes are same? A. \(^{81}C_2/^{100}C_2\) B. \(^{90}C_2/^{100}C_2\) C. \(^{162}C_2/^{100}C_2\) D. \(^{243}C_2/^{100}C_2\) E. \(^{270}C_2/^{100}C_2\) 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
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In a class of 100 students, all of them are coded with a distinct ....
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Updated on: 31 Jan 2019, 01:06
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 ; 100200,200300,300400,400500,500600,600700,700800,800900,9001000 27*9 = 243 P ()= 243c2/900c2 IMO D EgmatQuantExpert wrote: In a class of 100 students, all of them are coded with a distinct threedigit number from 100 to 199. If two students are select at random, what is the probability that exactly two digits of their codes are same? A. \(^{81}C_2/^{100}C_2\) B. \(^{90}C_2/^{100}C_2\) C. \(^{162}C_2/^{100}C_2\) D. \(^{243}C_2/^{100}C_2\) E. \(^{270}C_2/^{100}C_2\)
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Originally posted by Archit3110 on 30 Jan 2019, 05:58.
Last edited by Archit3110 on 31 Jan 2019, 01:06, edited 1 time in total.



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Re: In a class of 100 students, all of them are coded with a distinct ....
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30 Jan 2019, 10:49



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Re: In a class of 100 students, all of them are coded with a distinct ....
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30 Jan 2019, 19:46
EgmatQuantExpert wrote: In a school of 900 students, all of them are coded with a distinct threedigit 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. \(^{270}C_2/^{900}C_2\) 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.



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Re: In a class of 100 students, all of them are coded with a distinct ....
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30 Jan 2019, 23:40
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



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Re: In a class of 100 students, all of them are coded with a distinct ....
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31 Jan 2019, 13:26
kchen1994 wrote: EgmatQuantExpert wrote: In a school of 900 students, all of them are coded with a distinct threedigit 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. \(^{270}C_2/^{900}C_2\) 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. 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



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Re: In a class of 100 students, all of them are coded with a distinct ....
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01 Feb 2019, 01:02
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 3digit 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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Re: In a class of 100 students, all of them are coded with a distinct ....
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18 Feb 2019, 10:03
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 (09) 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




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