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How many fourdigit positive integers can be formed by using
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22 May 2012, 09:35
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How many fourdigit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ? A. 400 B. 1728 C. 108 D. 216 E. 432 I am getting answer as 432, but the OA is D. Here is the approach I used. First number can be selected from 1 to 9  in 9 ways, 2nd number has to be same as the one selected so only 1 way, Third number can be selected from remaining 8 numbers in 8 ways and 4th number again has to be only 1 so in total (9*1*8*1)*4!/2!*2! = 72*6 = 432 Can anyone please explain what mistake I am making here? Thanks!
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Re: How many fourdigit positive integers can be formed by using
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22 May 2012, 10:35
gmihir wrote: How many fourdigit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?
A. 400 B. 1728 C. 108 D. 216 E. 432
I am getting answer as 432, but the OA is D. Here is the approach I used.
First number can be selected from 1 to 9  in 9 ways, 2nd number has to be same as the one selected so only 1 way, Third number can be selected from remaining 8 numbers in 8 ways and 4th number again has to be only 1 so in total (9*1*8*1)*4!/2!*2! = 72*6 = 432 Can anyone please explain what mistake I am making here? Thanks! XXYY can be arranged in \(\frac{4!}{2!2!}=6\) ways (# of arrangements of 4 letters out of which 2 X's and 2 Y's are identical); # of ways we can select 2 distinct digits out of 9 is \(C^2_9=36\); Total # of integers that can be formed is 6*36=216. Answer: D.
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Re: How many fourdigit positive integers can be formed by using
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22 May 2012, 10:28
gmihir wrote: How many fourdigit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?
A. 400 B. 1728 C. 108 D. 216 E. 432
I am getting answer as 432, but the OA is D. Here is the approach I used.
First number can be selected from 1 to 9  in 9 ways, 2nd number has to be same as the one selected so only 1 way, Third number can be selected from remaining 8 numbers in 8 ways and 4th number again has to be only 1 so in total (9*1*8*1)*4!/2!*2! = 72*6 = 432 Can anyone please explain what mistake I am making here? Thanks! The number of ways u can chose two digits from 9 digits is 9C2=(9*8)/2. And the no of ways u can arrange two digits like asked is 4C2=(4*3)/2. total ways is 36*6=216. Hope that helps. If you like it give me Kudos




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Re: How many fourdigit positive integers can be formed by using
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22 May 2012, 10:29
This one is confusing to me. I know how to find the answer in a much more brute force manner.
Think about how many possible combinations you can make with 2 numbers (say 1 and 2)
1122, 1212, 1221, 2121, 2211, 2112 (6 combinations)
The number 1 has 8 numbers it can pair off with The number 2 has 7 numbers it can pair off with The number 3 has 6 numbers it can pair off with so on and so forth
\(= (8+7+6+5+4+3+2+1) * 6\)
\(= 36 * 6\)
\(= 216\)



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Re: How many fourdigit positive integers can be formed by using
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22 May 2012, 10:43
Cares wrote: This one is confusing to me. I know how to find the answer in a much more brute force manner.
Think about how many possible combinations you can make with 2 numbers (say 1 and 2)
1122, 1212, 1221, 2121, 2211, 2112 (6 combinations)
The number 1 has 8 numbers it can pair off with The number 2 has 7 numbers it can pair off with The number 3 has 6 numbers it can pair off with so on and so forth
\(= (8+7+6+5+4+3+2+1) * 6\)
\(= 36 * 6\)
\(= 216\) Regarding the pairing you are right.It is the basic way of understanding combinations of numbers. here you are picking 2 numbers to form a set from 9 available numbers,which turns out to be 9C2. Basically picking of k numbers from a n set of numbers gives nCk ways of possibilities. Hope its clear to you now.



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Re: How many fourdigit positive integers can be formed by using
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24 May 2012, 02:09
I understood the solution of Bunuel, but can you explain where did gmihir went wrong... Thanks in advance.



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Re: How many fourdigit positive integers can be formed by using
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24 May 2012, 02:16
pavanpuneet wrote: I understood the solution of Bunuel, but can you explain where did gmihir went wrong... Thanks in advance. In that solution the # of ways to choose 2 different numbers for X and Y is 9*8=72, but it should be \(C^2_9=36\), so gmihir's solution counts the same numbers twice.
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Re: How many fourdigit positive integers can be formed by using
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16 Sep 2012, 06:09
Bunuel wrote: gmihir wrote: How many fourdigit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?
A. 400 B. 1728 C. 108 D. 216 E. 432
I am getting answer as 432, but the OA is D. Here is the approach I used.
First number can be selected from 1 to 9  in 9 ways, 2nd number has to be same as the one selected so only 1 way, Third number can be selected from remaining 8 numbers in 8 ways and 4th number again has to be only 1 so in total (9*1*8*1)*4!/2!*2! = 72*6 = 432 Can anyone please explain what mistake I am making here? Thanks! XXYY can be arranged in \(\frac{4!}{2!2!}=6\) ways (# of arrangements of 4 letters out of which 2 X's and 2 Y's are identical); # of ways we can select 2 distinct digits out of 9 is \(C^2_9=36\); Total # of integers that can be formed is 6*36=216. Answer: D. Hi Bunuel, I Want to know if the statement changes from "How many fourdigit positive integers can be formed by using the digits from 1 to 9" to How many fourdigit positive integers can be formed by using the digits from 0 to 9", what would be answer. As per me answer should be 243. Kindly conform it.
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Re: How many fourdigit positive integers can be formed by using
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16 Sep 2012, 06:46
fameatop wrote: Bunuel wrote: gmihir wrote: How many fourdigit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?
A. 400 B. 1728 C. 108 D. 216 E. 432
I am getting answer as 432, but the OA is D. Here is the approach I used.
First number can be selected from 1 to 9  in 9 ways, 2nd number has to be same as the one selected so only 1 way, Third number can be selected from remaining 8 numbers in 8 ways and 4th number again has to be only 1 so in total (9*1*8*1)*4!/2!*2! = 72*6 = 432 Can anyone please explain what mistake I am making here? Thanks! XXYY can be arranged in \(\frac{4!}{2!2!}=6\) ways (# of arrangements of 4 letters out of which 2 X's and 2 Y's are identical); # of ways we can select 2 distinct digits out of 9 is \(C^2_9=36\); Total # of integers that can be formed is 6*36=216. Answer: D. Hi Bunuel, I Want to know if the statement changes from "How many fourdigit positive integers can be formed by using the digits from 1 to 9" to How many fourdigit positive integers can be formed by using the digits from 0 to 9", what would be answer. As per me answer should be 243. Kindly conform it. Yes, the answer should be 243. Assume the first digit is A, for which we have 9 choices, because it cannot be 0. Then, we have 3 choices where to place the second digit A. Now, for the other different digit B, we have again 9 choices (different from A but can be 0), and the places of the two B's are uniquely determined. Therefore, total number of possibilities is 9*3*9 = 243. A similar logic can be applied also for the original question: 9*3*8 = 216 9 possibilities for A 3 places to choose from for the second A 8 possibilites for B (cannot be A and cannot be 0), no places to choose, already determined
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Re: How many fourdigit positive integers can be formed by using
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18 Sep 2012, 23:07
we units 1,2,3,4,5,6,7,8,9 total 9 digits Now we want arragement xxyy x  can be selected in 9 ways x  same as first x  1 way y  can be selected from remaining 8 digits  8 ways y  same as first y  1 way so total# of ways = 9*1*8*1 = 72 not the 4 digits can be organised in 4C2 ways = 6 ways So total arragements = 72*6 = 432 Am I wrong in thinking? & I didnt understand the concept mentioned by Bunuel here... P & C is my weakest point in Quant



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Re: How many fourdigit positive integers can be formed by using
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19 Sep 2012, 00:00
bhavinshah5685 wrote: we units 1,2,3,4,5,6,7,8,9 total 9 digits Now we want arragement xxyy x  can be selected in 9 ways x  same as first x  1 way y  can be selected from remaining 8 digits  8 ways y  same as first y  1 way so total# of ways = 9*1*8*1 = 72 not the 4 digits can be organised in 4C2 ways = 6 ways So total arragements = 72*6 = 432 Am I wrong in thinking? & I didnt understand the concept mentioned by Bunuel here... P & C is my weakest point in Quant the 4 digits can be organised in 4C2 ways = 6 waysYou have to divide by 2, so only 3 ways. Or, 9 possibilities for the first digit, this can be placed for the second time in 3 places, then in the remaining places have another different digit  8 possibilities. A total of 9*3*8=216. 4C2 would work if you choose two distinct digit out of 9, which is 9C2 = 36 and you don't care about the order in which you have chosen them, take 2 of each type, then arrange them, which is 4C2 = 6. This will give you again 9C2*4C2 = 36*6 = 216. You cannot in one stage take order into account, then in the next stage ignore order. Choosing two digits as 9*8 means you distinguish between which one is chosen first. For example, once you considered x = 4 and y = 1, which you can arrange in 4C2 = 6 ways. But in this 6 arrangements are also include those with a 1 in the first place. Then you considered the pair x = 1 and y = 4, for which again 4C2 = 6 arrangements, but these are identical to the previous ones. 4C2 counts arrangements without distinguishing whether you first place x or y.
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Re: How many fourdigit positive integers can be formed by using
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24 Sep 2012, 03:42
Numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 ie 9 different numbers Lets assume we have a XXYY scenario , this means that we could have 9x1x8x1 = 72 different numbers We could also have XyXy combination so that is also 72 different numbers We could have a XYYX combination and that is 72 more different numbers ... 72 x 3 = 216... D is the answer ...
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How many fourdigit positive integers can be formed by using
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05 May 2016, 05:27
Stage 1 : Number of ways to select 2 different digits form 9 => (9 * 8 )/2 =36
Stage 2 : Number of ways to arrange AABB (MISSISSIPPI rule) = (4 * 3 *2) / (2*2) = 6
Fondamental Counting Principle = 36 * 6 = 216



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How many fourdigit positive integers can be formed by using
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Updated on: 01 Jun 2016, 12:33
surbhi87 wrote: yeah i did but don't quite understand the logic could you pls elaborate a little? If anyone else wanders into this forum and has the same question like Surbhi87 and I: I thought perhaps, the simple explanation for why the number of ways to pick to digits by simply multiplying 9 x 8 = 72 is because this is not a permutation, but a combination problem, where the order of selecting numbers does not matter. Am I correct?
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Last edited by fantaisie on 01 Jun 2016, 12:33, edited 1 time in total.



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Re: How many fourdigit positive integers can be formed by using
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01 Jun 2016, 09:21
fantaisie wrote: surbhi87 wrote: yeah i did but don't quite understand the logic could you pls elaborate a little? If anyone else wanders into this forum and has the same question like Surbhi87 and I: I though perhaps, the simple explanation for why the number of ways to pick to digits by simply multiplying 9 x 8 = 72 is because this is not a permutation, but a combination problem, where the order of selecting numbers does not matter. Am I correct? Exactly, you can have 1144, or 4141, or 4411, or etc... Since you have two pairs of equal digits, the order doesn't matter here.



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Re: How many fourdigit positive integers can be formed by using
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04 Jun 2016, 12:24
Bunuel wrote: pavanpuneet wrote: I understood the solution of Bunuel, but can you explain where did gmihir went wrong... Thanks in advance. In that solution the # of ways to choose 2 different numbers for X and Y is 9*8=72, but it should be \(C^2_9=36\), so gmihir's solution counts the same numbers twice. I always tend to miss the cases where a solution is counted twice. Could you please explain as to how i can get it right?



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Re: How many fourdigit positive integers can be formed by using
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26 Jun 2017, 17:54
gmihir wrote: How many fourdigit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?
A. 400 B. 1728 C. 108 D. 216 E. 432 Let’s first find the number of ways we can form a number AABB where A and B denote distinct digits. For the first digit, we have 9 possible options. If the second digit matches that of the first, then we have 1 option. Then for digit 3, we have 8 possible options. Since digit 4 matches digit 3, we have 1 option. So, for this scenario, we have 9 x 1 x 8 x 1 = 72 options. Since we are essentially being asked how many ways we can arrange As and Bs in AABB, where A and B are numbers 1 through 9, we see that AABB can be arranged in 4!/(2! x 2!) = 24/4 = 6 ways. Finally, we should take into account that we counted each number twice when we made this calculation (for instance, when A = 1 and B = 2, AABB = 1122, but when A = 2 and B = 1, BBAA = 1122); therefore, we should divide the result by 2. Thus, the total number of ways to create a 4digit number with 2 unique digits is (6 x 72)/2 = 216. Answer: D
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Re: How many fourdigit positive integers can be formed by using
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06 Jul 2017, 20:11
Of the threedigit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?
Then how come this question is not 9C2 * 3C2?



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Re: How many fourdigit positive integers can be formed by using
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16 Jul 2017, 04:09
Quote: the 4 digits can be organised in 4C2 ways = 6 ways You have to divide by 2, so only 3 ways. Or, 9 possibilities for the first digit, this can be placed for the second time in 3 places, then in the remaining places have another different digit  8 possibilities. A total of 9*3*8=216.
4C2 would work if you choose two distinct digit out of 9, which is 9C2 = 36 and you don't care about the order in which you have chosen them, take 2 of each type, then arrange them, which is 4C2 = 6. This will give you again 9C2*4C2 = 36*6 = 216.
You cannot in one stage take order into account, then in the next stage ignore order. Choosing two digits as 9*8 means you distinguish between which one is chosen first. For example, once you considered x = 4 and y = 1, which you can arrange in 4C2 = 6 ways. But in this 6 arrangements are also include those with a 1 in the first place. Then you considered the pair x = 1 and y = 4, for which again 4C2 = 6 arrangements, but these are identical to the previous ones. 4C2 counts arrangements without distinguishing whether you first place x or y. This is the best explanation of the conundrum by far. I was also confused at first, not understanding why the answer is 216, not 432. In fact, if we pick the 2 numbers as a first step using "9 x 8 = 72", we are picking for example 8 & 9 as well as 9 & 8. Which means we have already "arranged" the 2 numbers one time. And then when we arrange the 2 numbers using 4!/2!*2!, we are again arranging the same numbers, now for a second time. That is why double counting is involved. The way to solve this is to NOT arrange the numbers when we pick them in the first place, meaning we should use "combination" instead of "permutation". So, choose 2 numbers out of 9 (9C2 = 36) and then arrange those 2 numbers in 4!/2!*2! ways. Hope this helps



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How many fourdigit positive integers can be formed by using
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20 Jul 2018, 10:50
If I arrange two distinct things in four places  4!/2!*2! = 6 Select two from nine  9c2 So total  9*8*6/2 = 216




How many fourdigit positive integers can be formed by using
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