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HOT Competition 1 Sep/8AM: A locker code consists of five digits, for

1 2 3 4

Bunuel

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01 Sep 2020, 08:12

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Option A

5 digit codes - 10^5

5 digit codes with the condition that 9 immediately follows 5 are

Let's assume 59 as a single entity

Then 59 _ _ _ can be filled in 4* 10³ = 4000 ways.

The number of ways in which 9 does not follow 5 immediately are
10^5 - 4000
96000
Option A

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Total number of ways of generating code = 10^5

Total codes wherein 5 is followed by 9 = 4*10^3

Total codes wherein 5 is not followed by 9 = 10^5 - (4*10^3) = 96000

Option A should be the answer.

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total ways to arrange 5 digits is ; 10^5 and given condition that 5 is not followed by let 9
Let 95 be X so total places would be X10_10_10_ ;
we have 4*10^3 ways to put 5 next to 9 and to not put it this way total ways ;
10^5-4*10^3 ;
option A 96,000

A locker code consists of five digits, for example 04391, 23335, 05531... David wants to set his locker code so that digit 5 is not immediately followed by the digit 9. How many such codes are possible?

A. 96,000
B. 96,001
C. 96,002
D. 96,003
E. 96,004

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01 Sep 2020, 08:19

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5 digits can be placed with 10 numebrs ( 0,1,2,...9)
total possible digits = 100000

If 5 is immediately followed after 9 then total digits would be
10 * 10*10 *(9 ) (5)
(9)(5) can be at 1st digit, 2nd ,3rd or 4th
so total values in whih 95 together: 4*1000

without 5 followed after 9
= 10^5- 4000

96000- final answer (A)

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A locker code consists of five digits, for example 04391, 23335, 05531... David wants to set his locker code so that digit 5 is not immediately followed by the digit 9. How many such codes are possible?

A. 96,000
B. 96,001
C. 96,002
D. 96,003
E. 96,004

Since as per stated examples we can deduce that we can repeat the digits in 5 digit locker code

so for 5 position we have 10^5 options or 100,000

Now lets see codes with 95 together
95--- = 10^3
and since we can use 95 to 4 location in 5 digit locker code (95---,-95--,--95-,---95) total number of combination with 95 would be
4*10^3

so total number of locker code with no digit 5 followed by digit 9 =
10^5 - 4*10^3
=10^3(10^2-4)
= 96000

IMO A

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01 Sep 2020, 08:25

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5 digit code = 10^5
That 5 follows 9 we have [(10^3)*1]*4
Subtracting from 10^5. We get A

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First of all, let's consider the total no: of 5-digit codes that can be formed : P1 P2 P3 P4 P5 are the positions of the code.

P1 can be taken by any of the 10 digits from 0 to 9, since 0 is also allowed in P1.

Similarly P2, P3, P4, & P5 can also be taken by any of the 10 digits, since repetition of digits is allowed.

So, the total no: of 5-digit codes that can be formed is 10x10x10x10x10 = 10^5

Since David wants to set his locker code so that digit 5 is not immediately followed by the digit 9, let's remove all such codes, in which 5 is immediately followed by 9, from the total.

When 5 takes P1 & 9 takes P2, the remaining positions : P3,P4, & P5: can be taken by any of the 10 digits i.e. 10x10x10 = 10^3 codes

Similarly, When 5 takes P2 & 9 takes P3, we get 10 ^ 3 codes

When 5 takes P3 & 9 takes P4, we get 10 ^ 3 codes

When 5 takes P4 & 9 takes P5, we get 10 ^ 3 codes

So, the total no: of codes in which 5 is immediately followed by 9 = 4 * 10^3
= 4000

Since David wants to exclude all such codes : (Total no: of codes) - (no: of codes in which 5 is immediately followed by 9 )

=> 10^5 - 4000

=> 96000

Hence, A is the answer

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A locker code consists of five digits, for example 04391, 23335, 05531... David wants to set his locker code so that digit 5 is not immediately followed by the digit 9. How many such codes are possible?

A. 96,000
B. 96,001
C. 96,002
D. 96,003
E. 96,004

Total combinations: \(10^5\)
Number of Combinations in which 9 follows 5: \(4* 10^3\)
Required number of combinations: \(10^5 - 4* 10^3\)
96000
Answer A

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Total number of possible combinations = \(10*10*10*10*10 = 10^5\)

Possible combinations for 9 being followed by 5 -
95xxx - \(10*10*10 = 10^3\)
x95xx - \(10*10*10 = 10^3\)
xx95x - \(10*10*10 = 10^3\)
xxx95 - \(10*10*10 = 10^3\)

Total Combinations Possible as per the question = \(10^5 - 4*10^3 = 96000\)

A

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Total possibilities minus possibilities in which 59 are together in same order

Total possibilities are 10^5
Possibilities when 5 and 9 are together are 4C1 * 10^3

Solving we get 96000

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01 Sep 2020, 08:38

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The total number of possible codes is for a five digit number is - 10^5

9 and 5 should be together - so total number of possible codes is 4×10^3.

Hence answer is 10^5-4×10^3 = 96,000

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_ _ _ _ _
Total possibility = 10^5

(9,5) _ _ _
Possibility = 4C1 *10*10*10 = 4000
Required possibilities = 10^5 - 4000 = 96000
Hence A

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Given: If locker code consists of five digits, for example 04391, 23335, 05531... David wants to set his locker code so that digit 5 is not immediately followed by the digit 9.

Asked: How many such codes are possible?

Total codes consisting of 5 digits =10^5 =100000
Codes consisting of the sequence 59 = 10^3×4=4000
Since there are 10^3 ways to choose remaining 3 digits and 4 ways in which location of 59 may be chosen.

Total codes in which 5 is not immedately followed by 9 = 10000-4900 = 96000

IMO A

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01 Sep 2020, 08:47

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A locker code consists of five digits, for example 04391, 23335, 05531... David wants to set his locker code so that digit 5 is not immediately followed by the digit 9. How many such codes are possible?

since numbers can be repeated all cases will be \(10*10*10*10*10=10^5\)
cases where 5 is followed by 9 will be \(4*10^3=4000\)
\(5,9,10,10,10~10^3\)
\(10,5,9,10,10~10^3\)
\(10,10,5,9,10~10^3\)
\(10,10,10,5,9,~10^3\)

\(10^5-4000=96000\\
\)

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I am confused with number of cases I am getting in this question. Please help how to solve in a shorter way

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A locker code consists of five digits, for example 04391, 23335, 05531... David wants to set his locker code so that digit 5 is not immediately followed by the digit 9. How many such codes are possible?

First of all, total number of codes is 10^5
Now we will take [59] as one single number.
[59]XXX it has 1C4 different arrangement, which is 4. The other 3 digits can have 10^3 values
Thus 10^5-4*10^3=10000-4000=9600
The answer is A

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01 Sep 2020, 08:56

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Suppose that 5 follows 9
59_ _ _
_ 59_ _
_ _ 59_
_ _ _59
These will be the possibility that is 10 possible way of writing first _
So 10*10*10=1000*4 for four option ie 4000
Total possiblity is 10*10*10*10*10=100000-4000=96000 possibility

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A locker code consists of five digits, for example 04391, 23335, 05531... David wants to set his locker code so that digit 5 is not immediately followed by the digit 9. How many such codes are possible?

A. 96,000
B. 96,001
C. 96,002
D. 96,003
E. 96,004

Here we should try to find cases where 5 comes exactly after 9. That will be 4000 such cases. Total sample 100000 possibilities minus 4000 is 96000 possibilities.

So option A is the correct answer.

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A locker code consists of five digits, for example 04391, 23335, 05531... David wants to set his locker code so that digit 5 is not immediately followed by the digit 9. How many such codes are possible?

A. 96,000
B. 96,001
C. 96,002
D. 96,003
E. 96,004
----
Howdy -- Need to take 10^5 total possibilities and then figure out the number of possibilities that do not fit this. From here, we know that B is the correct answer.