Latest TCS Aptitude Question SOLUTION: 7. How many numbers greater than a lakh can be formed with the numbers 0, 1, 2, 3, 2,4

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7. How many numbers greater than a lakh can be formed with the numbers 0, 1, 2, 3, 2,4

Asked In TCS Bishal Pandey (10 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

Neha Rani  (10 years ago)

ans=((5*6*6*6*6*6)/2!) -1
from left 1st place can be filled in 5 ways excluding zero
rest five place can be filled in 6 ways. divide whole by 2! bcz we have two 2's which are identical.subtract 1 from result bcz we have only one case in which we get the no. equal to 100000 but we have to take more than 100000.

Is this solution Helpfull? Yes (34) | No (3)  

abhishek bathla  (10 years ago)

ans=6^5-1

all the boxes for 6 digit no. can be filled by 6 ways execpt 1st block it can be filled by 5 ways
so,
5*6*6*6*6*6
now 1 has to be subtracted of becoz from this there will be the case when 100000 occurs but we wnt more thn that.
so 5*6*6*6*6*6-1

Is this solution Helpfull? Yes (33) | No (8)  

Md Shahnawaz Shahab  (10 years ago)

5*6^5 - 1 = 5*7776 - 1 = 38880-1 =38879

Is this solution Helpfull? Yes (9) | No (2)  

Saloni Singh  (10 years ago)

5*6*6*6*6*6
as 1st place can b filled in 5 ways(xcluding 0)& the rest places can be filled in 6 ways each

Is this solution Helpfull? Yes (8) | No (4)  

shibam jha  (10 years ago)

Ans: infinite numbers.
because question is grater than a lakh but there s no limit upto which we have to find.. So simply infinite number can be made.
5*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6*6................. upto infinite nmbr..

Is this solution Helpfull? Yes (7) | No (8)  

Ankit Jajoo  (10 years ago)

without repetition
total numbers= 6!/2!=360
minus when 0 is in first place => 5!/2!=60

therefore total no. above lac= 360-60=300 Ans

Is this solution Helpfull? Yes (6) | No (2)  

Himanshu  (10 years ago)

ans=480
4*5*4*3*2*1

Is this solution Helpfull? Yes (3) | No (11)  

ankit  (10 years ago)

since 2 is given twice and so we hve only 5 different digits including 0.
first cant be 0 so only 4 ways to fill it and rest all by 5 ways

ans = 4*5*5*5*5*5-1
=12499

Is this solution Helpfull? Yes (2) | No   

Ritesh Anand  (10 years ago)

the answer will be 600,if repeatition is not allowed.

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

Adarsha dash  (10 years ago)

total numbers can be created - 6!/2!=360(two '2's are there)..
total number which is less than 100000 = 1*5*4*3*2*1/2!=60 (0 is selected first, then rest 5 nos can be arranged as 5!/2!)
so, numbers greater than a lakh = 360-60 =300.. for more tips n tricks add me on facebook, mah id-adi1nonly .. good luck

Is this solution Helpfull? Yes (1) | No   

anirban kundu  (10 years ago)

if repeatation is allowerd then...
4*5*5*5*5*4=10,000

Is this solution Helpfull? Yes | No (5)  

Arindam Dhara  (10 years ago)

for a number to be greater than lakh it should be of 6 digits.....1st digit can be any digit except 0...rest digits can be anyone from the given digits...
5*6*6*6*6*6=38880

Is this solution Helpfull? Yes | No (2)  

shubham agrawal  (9 years ago)

When we take 1,3,4 as lachth term then next 5 digits can arrenged in 5!/2! Ways bcoz 2 is repeated 2 times.
For 2 as lakhth term last 5 digits all are now different nd can be arrenged in 5! Ways.
0 cant be lakhth term so total ways are
3*5!/2!=180
5!=120
Total are 300

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Code Wizard  (5 years ago)

Total ways they can be arranged – without 0 at first they can be arranged
6!/2!=360 -- 5!/2!=60
300

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