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

1 2 3 4

TarunKumar1234

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01 Sep 2020, 08:58

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There are 5 places for digits to fill in.

In normal case= 10*10*10*10*10= 100000

Now, when 9,5 are fixed at 1st and 2nd place, we have combinations= 1000
when 9,5 are fixed at 2nd and 3rd place, we have combinations= 1000
when 9,5 are fixed at 3rd and 4th place, we have combinations= 1000
when 9,5 are fixed at 4th and 5th place, we have combinations= 1000

So, 5 immediately followed by 9 combinations= 4000

Not 5 immediately followed by 9 combinations= 100000-4000= 96000

So, I think, Ans. A.

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01 Sep 2020, 09:02

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Ans: 903

Total 5 digit combination = 10^5 = 100000
Now possible scenarios where 9 will be followed by 5: 59xxx, x59xx, xx59x, xxx59
Hence possible scenarios = 4*1000
But in the above combinations there will be repeat scenarios : 59x59, x5959, 59x59

So total possible solutions: 100000-4000+3
=96003

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01 Sep 2020, 09:10

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Quote:

Solution:

=> Total possible combinations (without any condition) = 10^5

=> Number of combinations where 5 is immediately followed by the digit 9.
=> 10^3 * 4 (4 possible cases, when 5 is in first, second, third and forth place.)

For, 5 is not immediately followed by the digit 9
= 10^5 - 10^3 * 4
= 10^3 (100 - 4)
= 1000 * 96
= 96,000

IMO answer is A

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01 Sep 2020, 09:20

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IMO A

Combinations are
(10!*9!*10!*!10!*10!) -> there can be 5 such combinations so (10!*9!*10!*!10!*10!)*5
Now as this number is multipied by 5 the only possible values at units place can be 0 or 5
Hence A

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01 Sep 2020, 09:22

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Total possible cases=

10x10x10x10x10=10⁵

(0 is also allowed at first place in case of passwords)

Now we have to calculate the cases in which 5 is followed by 9 immediately.

Assume (9,5) as one entity..
So total 4 places has to be filled with 10 digits from 0 to 9..

(9,5) can be places at any of the 4 places in 4C1 ways..
And the rest three can be filled in 10 ways each..so total such cases will be.

4C1x10x10x10=4000

Required number of cases will be 100x10³-4x10³=96x10³

ANS:A

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01 Sep 2020, 09:25

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The no of codes possible = total no of combinations- no of combinations when 5 is followed by 9
Total no of combinations = 10*10*10*10*10= 10^5
5 followed by 9 = taking 5&9 as one block
Hence now _ _ _ _ can be arranged in 4* 10*10*10= 4000

Hence total when 9 doesn’t follow 5 ==
100000 - 4000
96000

IMO A

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debjit1990

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01 Sep 2020, 09:28

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Total no of codes that can be made=10^5
total 5 spaces are there
let, 59 occupies the first two places..so other three places can be filled with 10^3
now this 59 can occupy a total of 4 places
so with 59 the total no of codes=4*10^3
withoud 59 following=10^5-(4*10^3)=96000
ans :A

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01 Sep 2020, 09:31

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Quote:

Total possibilities (without any constraints) = 10x10x10x10x10

Number of ways 59 can come together and in that order
5 9 _ _ _
_ 5 9 _ _
_ _ 5 9 _
_ _ _ 5 9
Thus, 4 ways
In each case, each of the other 3 spots can be filled in 10*10*10 ways
Thus 4000 total cases

We need to remove these cases from the total number of cases to get our answer

=> answer = 10^5 - 4*10^3 = 96000
Answer: A

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01 Sep 2020, 09:32

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IMO A

5----
-5---
--5--
---5-
----5

5 can take this places and not followed by 9.
5's position is fixed.
Next blank can take any of (012345678)values so 9!
other blanks can take 10!
10!×9!×10!×10!×5
We multiply by 5 because of 5 different placements of 5
So answer should be multiple of 5
Only A is.

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01 Sep 2020, 09:33

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total combinations - 10^5
with condition - 5 at 10^5 place 9 at 10^4 place - so 1000 numbers like these
likewise keep on rolling 5 and 9 position -- total 4000 numbers - 10^5 - 4000 so 96000

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01 Sep 2020, 09:34

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Total no. of codes possible will be 10*10*10*10*10 = 10^5

Cases in which 5 will be followed by 9,
5-9-x-x-x --> 1*1*10*10*10
x-5-9-x-x --> 10*1*1*10*10
x-x-5-9-x --> 10*10*1*1*10
x-x-x-5-9 --> 10*10*10*1*1

So total no. of such cases will be (10*10*10) * 4 = 4000

But there will be 3 cases in which "5-9" will repeat in a code
5-9-5-9-x
x-5-9-5-9
5-9-x-5-9
These cases are counted twice in the above , so we will subtract these 3 cases from 4000

Hence total no. of ways = 10^5 - 3997 = 96003

Correct answer is D

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01 Sep 2020, 09:38

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Total number of cases (without any restriction) of code consists of five digits = 10*10*10*10*10 = 100000

Now,

cases in which digit 5 is immediately followed by the digit 9 are :

5 9 _ _ _
_ 5 9 _ _
_ _ 5 9 _
_ _ _ 5 9

In each of above cases, number of ways = 10*10*10

Therefore total ways in which digit 5 is immediately followed by the digit 9 = 4 * 10*10*10 = 4000

Therefore,

Total ways in which digit 5 is NOT immediately followed by the digit 9 = 100000 - 4000 = 96000

Hence, Correct answer is Option A.

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01 Sep 2020, 09:40

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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?

Total 5 slots _ _ _ _ _ needs to be filled with 10 digits.
Total no of possibilities is 10X10X10X10X10
5 must not be followed by 9.
5 9 _ _ _ : 10X10X10 = 1000 ways
_ 5 9 _ _ : 10X10X10 = 1000 ways
_ _ 5 9 _ : 10X10X10 = 1000 ways
_ _ _ 5 9 : 10X10X10 = 1000 ways
4000 ways in which 5 will be followed by 9.
Subtract this from the total no of ways. Option a will be the answer .

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01 Sep 2020, 10:19

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We can find this by first finding the total number of 5 digit codes and then subtracting the codes in which 5 is immediately followed by 9.

Total 5 digit codes
= 10*10*10*10*10
= 100000

To find codes in which 9 follows immediately after 5->
We have 4 cases:
a) 5 is at 1st position and 9 at 2nd
5 9 _ _ _
We can have 10*10*10 codes which start in this way

b) 5 is at 2nd position
And 9 at 3rd
_ 5 9 _ _ _
Again other positions can be filled as 10*10*10

c) 5 is at 3rd position and 9 at 4th
_ _ 5 9 _
Total codes = 10*10*10

d) 5 is at 4th position
And 9 at 5th
_ _ _ 5 9
Total codes 10*10*10

We can total these cases to be 4*10*10*10
= 4000

Let's subtract from total cases->
100000-4000
= 96,000

IMO A

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01 Sep 2020, 10:29

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The answer is D

Given that :
Code of 5 digits from 10 digits available ie.0 to 9

Condition:
5 not immediately followed by 9.

Total number of locks possible = 10×10×10×10×10 = 10^5

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01 Sep 2020, 10:36

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Number of arrangements where 5 is not followed by 9 is =
Total arrangements - Number of arrangements where 9 and 5 are together

Total arrangements = 10^5 = 100000

Number of arrangements where 9 and 5 are together = 3999

Thus answer is 96001.

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01 Sep 2020, 10:42

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for codes with 5 not followed immediately by 9 we will subtract codes with 5 followed by 9 from total possible codes.

total possible codes as repetition is allowed = 10^5

codes that have 5 immediately followed by 9 are
take 59 as one unit remaining 3 places have 10 options each so total case 4×10^3 but in this there are 3 cases that have been repeated i.e. 5959x,59x59,x5959
hence total cases are 4×10^3 - 3

so 10^5-(4×10^3 -3)= 96003

option D is correct

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01 Sep 2020, 11:19

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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

Solution:

Without any condition, the total cases are:

1st , 2nd, 3rd, 4th and 5th digits can each take 10 cases : so total 10,000 cases

now, lets look at the restriction

5 9_ _ _ : No of cases for this condition is 10 X 10 X 10 = 1000
but four such cases are possible : 5 9_ _ _, _ 5 9_ _ , _ _ 5 9_ , and _ _ _ 5 9

So subtract 4 * 1000 from 10,000 cases

Answer = 96,000

Option A is the correct answer

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01 Sep 2020, 11:33

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A?

Total for 5 numbers would be 100000 possible.

Numbers in which 5 is followed by 9 ie 59 occur together are 59xxx, x59xx, xx59x,xxx59 ie 4 possible . Also note x positions can be filled in 10 ways. ===> 4000 numbers.

Thus answer = 100000 - 4000 = 96000

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01 Sep 2020, 11:50

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Is it A

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?

Total - restriction (combination 5 ,9 )
10^5 - 4 * 10^3 = 96000

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