Latest Elitmus Aptitude Question SOLUTION: From digit 0-9, how many no. can be made such that two adjacent digits are same. like-119,911.....

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(#M40032533) ELITMUS Numerical Ability question Apptitude Keep an EYE

From digit 0-9, how many no. can be made such that two adjacent digits are same. like-119,911.....

Asked In Elitmus aditya sharma (10 years ago)

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Mainak Das  (10 years ago)

112,113-119=9
220,221,223-229=9
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990-998=9 so total 9*9=81(consecutive digits at fast)
122,133,144,155,166,177,188,199=8
211,233,244,255,266,277,288,299=8
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911,922,933,944,955,966,977,988=8 so total 8*9=72(consecutive digits at last)
Now 111,222,333,444,555,666,777,888,999 so total=9(all digits same)
Again 100,200,300,400,500,600,700,800,900 so total=9(consecutive two zeroes at last)

So al total two consecutive digits same= 81+72+9+9=171(Ans)

Is this solution Helpfull? Yes (36) | No (2)  

Vivek Jain  (10 years ago)

11X & X11 = 10 + 9 = 19
similarly 22X & X22 .. 33X & X33 upto 99X & X99
So,(10+9)*9=171

Is this solution Helpfull? Yes (14) | No (1)  

ankitbalyan  (10 years ago)

i think 171 is the ans

Is this solution Helpfull? Yes (8) | No (3)  

vinod  (10 years ago)

let us consider three digit no --- say first two filled with 11 and third can be filled in 10 ways so ... similarly with starting 22 can be filled in 10 ways..
it gives 9*10=90 ways

then other possible is --- filling the 2nd and 3rd place with same no this can be done in 10 ways and first place with 9 ways because 0 we can place at hundred place.. it gives 9*10=90ways
so total =90+90=180ways but we want to subtract 9 ways because in 1st and 2nd possible we counted the numbers 111 222 333 444 555 666 777 888 999 two times so we are going to subtract 9 ways ....
final total no of ways are 180-9=171 ways

Is this solution Helpfull? Yes (6) | No (1)  

Avigyan  (10 years ago)

if only two adjacent digits need to be same then.. ans is 162.
if all three digits can be same.. then ans is 171

Is this solution Helpfull? Yes (4) | No   

Gaurav Malhotra  (10 years ago)

@ANKITBALYAN ans ki maa ki aakh , Solution nahi aata kya.

Is this solution Helpfull? Yes (4) | No   

RAKESH  (10 years ago)

no. containing 2 1's
11(2-9)=16 [two 1's is fixed & 3rd digit may be from 2-9]
similarly for 22,33,..., 99 for each 16 no. will be formed

no. containing 00 => 9(100,200,...,900)

total=9*16+9=153

Is this solution Helpfull? Yes (3) | No (12)  

deepak  (9 years ago)

@ rakesh u have taken 11(2-9)...and what about 110,220,330.....

Is this solution Helpfull? Yes (1) | No   

Anish Kumar  (9 years ago)

let the no. is abc
1st find unfavourable no.
a=1,2,3.......,9=9 possibilities(except 0)
b=9 possibilities(except a)
c=9 possibilities(except b)
Thus, possible unfavourable no.= 9*9*9=729
Total no=900
Thus, Favourable no=900-729=171

Is this solution Helpfull? Yes (1) | No   

swarup  (8 years ago)

qustn should have been given clearly like 2 digit number or 3 digit number or 4 digit number...

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Mainak Das  (10 years ago)

then what about 111,222,333,444....999?? these nos are also there.

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DIVYA V M  (10 years ago)

Can someone pls explain how its 171..

Is this solution Helpfull? Yes | No (2)  

anji  (10 years ago)

110-119=>10,211-911=>9 So for each digit there will be 19 possibilities

so 19*9=161;

& for digit 0=> 0 cant be in 100s position,so only 10 combinations will be possible. totally 161+10:171 combinations are possible

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