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How many different 4-digit numbers

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chetan2u
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How many different 4-digit numbers [#permalink] New post   13 Mar 2016, 05:43
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How many different 4-digit numbers can be written using the digits 1 to 8 without repetition such that the number always contains the digit 2?
A. 360
B. 560
C. 760
D. 840
E. 1260
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Official Answer
D
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sowragu
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Re: How many different 4-digit numbers [#permalink] New post   13 Mar 2016, 05:52
chetan2u wrote:
How many different 4-digit numbers can be written using the digits 1 to 8 without repetition such that the number always contains the digit 2?
A. 360
B. 560
C. 760
D. 840
E. 1260


OA after 3 days



The most restrictive case. We should have a two.
No repetition. Numbers range from 1 to 8.
= 1*7*6*5 = 210 ways we can write if '2' is in the first position.
= 4*(210)
=840 ways.

Another method.

Without restriction for 8 digits- 8*7*6*5 = 1680 Ways we can write.
Without restriction for 7 digits- 7*6*5*4 = 840 Ways.

With restriction = 1680 - 840 = 840 Ways.

IMO.D.
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Re: How many different 4-digit numbers [#permalink] New post   13 Mar 2016, 07:41
chetan2u wrote:
How many different 4-digit numbers can be written using the digits 1 to 8 without repetition such that the number always contains the digit 2?
A. 360
B. 560
C. 760
D. 840
E. 1260


OA after 3 days


All possibel combinations - combinations without using digit 2

(8*7*6*5) - (7*6*5*4) = 840
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Re: How many different 4-digit numbers [#permalink] New post   15 Apr 2017, 03:29
chetan2u wrote:
How many different 4-digit numbers can be written using the digits 1 to 8 without repetition such that the number always contains the digit 2?
A. 360
B. 560
C. 760
D. 840
E. 1260


OA after 3 days


Let's list all possibilities and then subtract those without 2.
\((8*7*6*5)-(7*6*5*4)\)
\(7*6*5*4=840\)

D.
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Re: How many different 4-digit numbers [#permalink] New post   15 Apr 2017, 04:09
4*7p3=840 and that implies option ->D
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Re: How many different 4-digit numbers [#permalink] New post   23 Nov 2018, 15:23
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the question is asking for what are some possible way to rearrange 7 numbers into 3 columns, since 2 always occupy 1 of 4 total columns,

it is permutation, not combination where the question would give you if there are 4 columns, what are some possible ways to choose 4 numbers from 1 to 8

since it is permutation, it is 7p3 for each column, 7p3 = 7! / (7-3)! = 7 * 6 * 5 = 210
now we have the permutation for each column, now, there is digit 2 which can occupy any of 4 of 4 column
so, multiply 4 withe the permutation for each column, answer = 4 * 210 = 840
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Re: How many different 4-digit numbers [#permalink] New post   28 Jun 2021, 08:35
First we fulfil the condition by selecting the digit 2 out of the given 8 digits = 1C1
Then we can choose any three digit from the remaining 7 digits = 7C3
These 4 digits can be arranged in 4! ways

Hence,
1C1 * 7C3 * 4! = 840 (D)
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Re: How many different 4-digit numbers [#permalink]
28 Jun 2021, 08:35
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