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How many four-digit positive integers can be formed by using

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Re: How many four-digit positive integers can be formed by using  [#permalink]

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New post 05 Aug 2018, 09:29
Explanation:

Two digits are equal to each other, and the remaining two are also equal to each other but different from the other two. So, the number may be of the form AABB or ABAB or BBAA, etc.

First we select 2 different digits A and B out of digits from 1 to 9
⇒ The 2 different digits can be chosen in 9C2 = 36 ways

Now, from the selected 2 digits, we are making 4 digits by repeating each digit once. These digits can be arranged at 4 places in 4!2!2! = 6 ways
(n things, of which p, q and r are alike, can be arranged in n!p!q!r!).

⇒ Required number of ways = 36 × 6 = 216.

Answer: D.

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How many four-digit positive integers can be formed by using  [#permalink]

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New post 24 Aug 2019, 02:11
gmihir wrote:
How many four-digit 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!


Asked: How many four-digit 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 ?

Number of 4-digit such numbers = \(^9C_2 * 4!/2!2! = 36 * 6 = 216\)

IMO D
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Re: How many four-digit positive integers can be formed by using  [#permalink]

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New post 24 Aug 2019, 04:29
4 digit no ; aabb
4!/2!*2! ; 6
and the total digits can be ; 2 digit to choose from 9 digits ; 9c2 ; 36
so total 4 digit + integes ; 36* 6 ; 216
IMO D

gmihir wrote:
How many four-digit 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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How many four-digit positive integers can be formed by using  [#permalink]

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New post 14 Sep 2019, 11:12
Bunuel wrote:
gmihir wrote:
How many four-digit 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.


Wouldn't the 4!/2!×2! Also select xyxy which is not to be selected?
Bunuel?

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How many four-digit positive integers can be formed by using   [#permalink] 14 Sep 2019, 11:12

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