How many 6 digit numbers are there which contains 3 9s?
Answers
How many 6-digit numbers contain exactly 4 different digits ?
My Solution is : There are 6 digits and 4 needs to be unique so either 2 digits can be same or 3 can be same.
Ans: Solution 1: (9C1*9C1*8C1*7C1*(4C1*3C1) * 6!/2!*2! + 9C1*9C1*8C1*7C1*(4C1*1C1)* 6!/3!). This should be the solution. But it gives wrong answer
But there is another solution mentioned which gives the correct answer.
Solution 2: This is the solution given online. C(10, 4)* [C(4, 2)* 6!/(2!)2 + C(4, 1)* 6!/3!] =10*9*8*7*5*13 Removing the cases starting with zero
= 9*9*8*7*5*13 = 294840 .
Both these approaches gives 2 different answer, now which one to use, I select 2 digits from 4 numbers should i use 4C2 or 4C1*3C1 , similarly selecting 4 digits from 10 number, is it 10C4(replace 10 with 9 in the expression later on) or is it 9C1*9C1*8C1*7C1.
Answer:
Or 6=(10–5)+1=5+1=6
THus we can say thetotal numbers from A to B are
N=(B-A)+1………………………..(1)
Thus total numbers from 100000 to 999999
Putting A=100000 and B= 99999 in(1)
N=(999999–100000)+1
=1000000–100000
=Smallest 7 digit No- Smallest 6 digit no
N=9,00,000
=9*100000
N=9*smallest 5 digit number
Thus we can write
Sum of n digit numbers=9*(smallest n-1 digit number)
N=9*10^(n-1)