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5 digit no can be arranged in 9*10*10*10*10 ways ( if repetition allowed )

3 2digit no can be inserted in _ _ _ _ _ 5 digit ID no in 2 ways

(i) ID starting with 2 so thr are 5 ways in whch 3-2's can be placed in 5 slots 222--,22--2,2--22,2-2-2,2-22- remaining 2 digits can be filled in 9ways ( 0-9 digits excluding 2) each, and they can be arranged among themselves in 2 ways so total 9*9*5*2 ways

(ii) ID not starting with 2 ,, thr are 4 ways -222-,-22-2,-2-22-,--222 1st digit can be filled by any digit 1-9 excluding 2, so 8 ways 5th slot can be filled by 9 ways n both themselves can be arranged among themselves in 2 ways so total 8*9*4*2

that brings us to (9*9*5*2 + 8*9*4*2)/9*10*10*10*10 = 9*2(45+32)/9*10*10*10*10 = 2*77/10000 = 1.52%

5 digit no can be arranged in 9*10*10*10*10 ways ( if repetition allowed )

3 2digit no can be inserted in _ _ _ _ _ 5 digit ID no in 2 ways

(i) ID starting with 2 so thr are 5 ways in whch 3-2's can be placed in 5 slots 222--,22--2,2--22,2-2-2,2-22- remaining 2 digits can be filled in 9ways ( 0-9 digits excluding 2) each, and they can be arranged among themselves in 2 ways so total 9*9*5*2 ways

(ii) ID not starting with 2 ,, thr are 4 ways -222-,-22-2,-2-22-,--222 1st digit can be filled by any digit 1-9 excluding 2, so 8 ways 5th slot can be filled by 9 ways n both themselves can be arranged among themselves in 2 ways so total 8*9*4*2

that brings us to (9*9*5*2 + 8*9*4*2)/9*10*10*10*10 = 9*2(45+32)/9*10*10*10*10 = 2*77/10000 = 1.52%

I hope the logic works out to be correct

First: there is some problems in calculations, then the answer doesn't match with the choices provided. _________________

I have been trying this problem for sometime and here is my approach.. Total number of 5 digit numbers = 99999-10000+1=90,000 No. of ways of arranging 3 digits as 2 = 5C3=10.

Arrangements are as follows. Here X can be any digit other than 2

I have been trying this problem for sometime and here is my approach.. Total number of 5 digit numbers = 99999-10000+1=90,000 No. of ways of arranging 3 digits as 2 = 5C3=10.

Arrangements are as follows. Here X can be any digit other than 2

Here the first X cannot be digit '0' so no. of arrangements with first digit as 'X' = 8*9=72. Total such arrangements = 72*5=360

no. of arrangements with first digit as '2'=9*9=81 Total such arrangements = 81*5=405

Now there is some double counting where 'XX' are together.I have counted 00,11,33,44,55,66,77,88,99 as different numbers.. So need to subtract these..

In combination XX222 '00' is not present. So 8 combinations In combinations '222XX', '22XX2', '2XX22', there are 3*9=27 combinations.

Hence 765-(27+8)= 730 Total probability= (730/90,000)*100 ~ 0.81%. IMO answer is A

Couldn't understand this part:

Quote:

Now there is some double counting where 'XX' are together.I have counted 00,11,33,44,55,66,77,88,99 as different numbers.. So need to subtract these..

In combination XX222 '00' is not present. So 8 combinations In combinations '222XX', '22XX2', '2XX22', there are 3*9=27 combinations.

What exactly do you mean by double counting. Request you to explain this in detail.... Probably would be better if you could use some numbers as example and explain.

Thanks, JT _________________

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|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Ok let me try with an example..Though not sure how to explain more clearly.. For the combination, say 2XX22, numbers can vary between be 20022 to 29922. Hence total of 29922-20022+1=100 numbers. Of these 100 numbers, 19 numbers have more than three 2s. Hence 100-19=81.

Of these 81 numbers the following numbers are counted twice..Here when the 100th digit and 1000th digit are same, the numbers are counted twice though infact they are the same... 20022,21122,23322,24422,25522,26622,27722,28822,29922 Hence need to subtract 9.

This happens again with 22XX2, 9 times as well.. 22002,22112,22332,22442,22552,22662,22772,22882,22992...

For XX222, only 8 such combinations exist.. 11222,33222,44222,55222,66222,77222,88222,99222. Here key is to look for unique combinations..

I hope this clarifies your question..And maybe there is a simpler method to solve this

Note: Although you came up with 10 ways, your listed 10 ways had an issue. Please check with the combination in ur post.

Considering the first 6 ways, which have 2 in as the first start digit (from left): In 222XX, the first X from left can be filled up in 9 different ways (total 10 digits and we exclude digit 2 from this. Hence 9 digits) Similarly the second X from left can also be filled up in 9 different ways. Reason Both X can be same also as there isn't any condition on it. Therefore 222XX can be filled up in 9X9 ways i.e. 81. (I guess u got his number with a different approach which is also correct. I have used the generic approach followed in probability! But no worries as we both are on the same page ultimately)

Now comes the substraction which I do not understand the reason for! To make my point clear, I would list the 222XX combinations for you:

In the above table I dont see any number which is repeated or duplicated as per you. Hence this is where I am lost!

Anyway.. continuing the above approach... in the first 6 ways, we can have 81 X 6 = 486 numbers!

For the remaining 4 ways, we have the following explanation: 7. XX222 8. X222X 9. X2X22 10.X22X2

In the above sequence of numbers, the first X from left can be filled up by 8 different digits. (Out of the 10 digits, we remove 2 and 0 leaving only 8 possible digits) The second X can anyway be filled up by 9 different digits (Leaving only 2 out of the 10 digits)

Hence for each way we have 8X9 = 72 numbers. Therefore for 4 ways we have 72X4 = 288 numbers! Summing the 10 ways, we get 486 + 288 = 774 numbers Total number possible = 9 X 10 X 10 X 10 X 10 = 90000 numbers

Therefore probability % = (774 / 90000) X 100 = 0.86%. I know the answer isn't among the options and this makes me feel a bit less confident. Do let me know if you identify any loops in my approach!

For the additional question stated: Find the probability that any three digits (exactly three) are the same. the approach is as follows:

Let us consider the following combinations. The combination would be same as consider for 2. Only difference is that we can replace 2 with Y for now.

Considering the first 6 ways: YYYXX - We can choose first Y in 9 different ways (Leave out 0 from 10 digits). Once first Y is selected, the other Ys would be the same and hence no selection required. The first X can be selected in 9 different ways (leave out the digit selected for Y from 10 digits). The second X can be selected in 8 ways (leave out the digit selected for Y and for the first X from 10 digits). Therefore we have YYYXX as 9 x 1 x 1 x 9 x 8 = 648 numbers Hence for 6 ways we have 648 x 6 = 3888 numbers

Considering the next 4 ways: XXYYY - We can chose the first Y in 10 different ways. As before, once first Y is selected, the other Ys would be the same and hence no selection required. Returning to X, the first X can be filled in 8 ways (leave out 0 and Y digit from 10 digits) and the second X can be selected in 8 ways (leave out the digit selected for Y and for the first X from 10 digits). Therefore we have XXYYY as 8 x 8 x 10 x 1 x 1 = 880 numbers Hence for the 4 ways we have 880 x 4 = 3520 numbers Summing the numbers possible in 10 ways = 3888 + 3520 = 7408 Total Numbers = 90000 (as before)

Therefore probability = 7408/90000 = 0.0823 = 8.23%

Please feel free to share your analysis...

Thanks, JT _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

hi i think it does not req such long calculations my ans A... there are 5 digits each place can have any of ten digits so total posb=10^5... now three places only two can be there and rest two can have any of remaining nine digits... so posb=1*1*1*9*9*5c3.. prob=.81%... although i might be still missing something.. ill try again

hi i think it does not req such long calculations my ans A... there are 5 digits each place can have any of ten digits so total posb=10^5... now three places only two can be there and rest two can have any of remaining nine digits... so posb=1*1*1*9*9*5c3.. prob=.81%... although i might be still missing something.. ill try again

Chetan, I guess this is wat you r missing. Please check my explanation below:

First of all we would need to consider total number of 5 digit IDs as : 9 x 10 x 10 x 10 x 10 = 90000 and not 10^5 as we cannot have 0 in the first place, else it would become 4 digit ID. Now as you consider for 3 places to be filled with digit 2, the possible number of combinations = 1 x 1 x 1 x 9 x 9 x 5c3 = 810 number. Again this includes the possibility of having digit 0 as the first or having digit 0 as both first and second. We need to subtract the number falling into this category from 810. Hence we again arrive at the 4 combinations as follows where 0 could be the first or second: XX222 X222X X2X22 X22X2 Its only in them we can have 0 as first. If u notice, we could have 0 as both first and second in the first combination only. For 0 as the first digit, we have the following posb = 1 x 1 x 1 x 1 x 9 x 4 = 36 For 0 as both as first and second digit we have only 1 posb i.e. 00222 or 222. Therefore we need to subtract 36+1 = 37 from 810. This gives us the actuall posb = 810-37 = 773

Hence probability % = \frac{773}{90000} x 100 = 0.858% \approx 0.86%

I know I do not match the options given but am a little lost as I feel the logic to be correct! I hope am on the right track!

Thanks, JT _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

hi i think it does not req such long calculations my ans A... there are 5 digits each place can have any of ten digits so total posb=10^5... now three places only two can be there and rest two can have any of remaining nine digits... so posb=1*1*1*9*9*5c3.. prob=.81%... although i might be still missing something.. ill try again

Chetan, I guess this is wat you r missing. Please check my explanation below:

First of all we would need to consider total number of 5 digit IDs as : 9 x 10 x 10 x 10 x 10 = 90000 and not 10^5 as we cannot have 0 in the first place, else it would become 4 digit ID. Now as you consider for 3 places to be filled with digit 2, the possible number of combinations = 1 x 1 x 1 x 9 x 9 x 5c3 = 810 number. Again this includes the possibility of having digit 0 as the first or having digit 0 as both first and second. We need to subtract the number falling into this category from 810. Hence we again arrive at the 4 combinations as follows where 0 could be the first or second: XX222 X222X X2X22 X22X2 Its only in them we can have 0 as first. If u notice, we could have 0 as both first and second in the first combination only. For 0 as the first digit, we have the following posb = 1 x 1 x 1 x 1 x 9 x 4 = 36 For 0 as both as first and second digit we have only 1 posb i.e. 00222 or 222. Therefore we need to subtract 36+1 = 37 from 810. This gives us the actuall posb = 810-37 = 773

Hence probability % = \frac{773}{90000} x 100 = 0.858% \approx 0.86%

I know I do not match the options given but am a little lost as I feel the logic to be correct! I hope am on the right track!

Thanks, JT

No, that was not the thing chetan2u was missing. First ID number CAN start with 0, no problem with that. It's 5 digit number which can not start with 0. Can't you have ID (credit card) with number 01234?

In that case, I guess chetan2u was correct. Total posb = 10^5 Posb of having 3 2 digits = 1 x 1 x 1 x 9 x 9 x 5c3 = 810 Therefore probablity % = \frac{810}{10^5} x 100 = 0.81 %

Guess would wait for your solution!

Cheers! JT _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

I was beginning to rethink whether I should restart my Math prep again

Thanks for clarifying the same Bunuel!

Cheers! JT _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

My approach was finding total possibilities, and dividing out the number of possibilities with exactly three 2s.

Total possibilities = 100000

Two of the digits have 9 possible choices each, and the other three must be 2s, and the number of ways of choosing the three spaces that must have 2s is 5c3 = 10

Uhm, what does ID mean? My approach: there are 2 types of number satisfying: 2abcd (2 of a,b,c,d are digit 2) and Amnpq (A is different from 2; 3 of m,n,p,q are digit 2) 2abcd: 4C2 * 9 * 9 desired outcomes Amnpq: 8* 4C3 *9

Found this question elsewhere and think it was solved incorrectly. Give it a try. Only my solution to follow as no OA.

1) In a 5 digit ID number, what is the probability of exactly three digits are the digit 2? 2) Additional question: Find the probability that any three digits (exactly three) are the same.

1) total ways = 10^5 3 digits of 2 and the rest 2 digits are different = 9 * 9 * 5c3 (5c3 to pick 3 places for 2's and the rest possibilities are filled with one of the remaining 9 numbers) probability = 0.81% 2) extension to the above problem - multiply by 10 as there are 10 single digit numbers probability = 8.1%

1x1x1x9x9 = 81. This much is understood. I also understand that there are 5C3 ways in which the number 2 can be distributed among the 5 places. But what happens when the other 2 places are filled by the same number? For example, if the ID has the digits 22255 then these can form 5!/3!x2! numbers i.e. 10 numbers. But if the ID has 22237, for example, then there are 5!/3! = 20 combinations. Therefore, we cannot simply perform the multiplication of 81 with 5C3. In fact, we have to treat these cases separately as follows: 1x1x1x9x8 x 5!/3! = 1440 (here the other two numbers are different) 1x1x1x9x1 x 5!/3!x2!= 90 (here the other two numbers are the same) Therefore, the total possibilities are 1440+90=1530. Hence the prob. is 1530/90,000= .017%

The question is ambiguous. No where it is mentioned that 00000 could be taken as a 5 digit ID number. Going by actual 5 digit no.s, I think the answer should be 0.86 %

No. of 5 digit numbers = 9*10*10*10*10 = 90,000 Possible no.s with 2 in the ten thousands position = 1*4C2*9*9= 486 Possible no.s with any other no. in the ten thousands position = 8*4C3*9 = 288