1. ## [SOLVED] permutations 2

Find the number of 6 digit numbers that can be formed using the digits 0 to 9 if:

a) each digit number be different
b) digits may be repeated
c) at least one 5 is included
d) the number is even,greater than or equal to 600,000, with repetition of digits allowed

2. Hello, william!

I will assume that the number may not begin with 0 (zero).

Find the number of 6-digit numbers that can be formed using the digits 0 to 9 if:

a) each digit must be different
The first digit cannot be zero: 9 choices.

The second digit can be any of the other 9 other digits.
The third digit can be any of the other 8 digits.
The fourth digit can be any of the other 7 digits.
The fifth digit can be any of the other 6 digits.
The sixth digit can be any of the other 5 digits.

Answer: .$\displaystyle 9 \times 9 \times 8 \times 7 \times 6 \times 5 \:=\:130,\!080$

b) digits may be repeated
The first digit has 9 choices.
Each of the other five digits has 10 choices.

Answer: .$\displaystyle 9 \times 10^5 \:=\:900,\!000$

c) at least one 5 is included
In part (b), we saw that there are $\displaystyle 900,\!000$ possible numbers.

Consider the numbers that have no 5's.

The first digit has 8 choices (not 0 and not 5).
Each of the other five digits has 9 choices.
Hence, there are: .$\displaystyle 8 \times 9^5 \:=\:472,392$ numbers with no 5's.

Answer: .$\displaystyle 900,\!000 - 472,\!392 \:=\:427,608$

d) the number is even, greater than or equal to 600,000, with repetition allowed
The number is $\displaystyle \geq 600,\!000$
The first digit must be: 6, 7, 8, or 9 . . . 4 choices.

The number is even.
The last digit must be: 0, 2, 4, 6, 8 . . . 5 choices.

Each of the other four digits has 10 choices.

Answer: .$\displaystyle 4 \times 5 \times 10^4 \:=\:200,\!000$

3. Originally Posted by william
Find the number of 6 digit numbers that can be formed using the digits 0 to 9 if:

a) each digit number be different
You can choose any of the 10 digits to be first, any of the remaining 9 digits to be next, any of the remaining 8 digits to be next, ..., down to any of the remaining 5 remaining digits to be the last, 6th digits. There are 10*9*8*7*6*5= 10!/(10-4)!= 9!/4! such digits.

b) digits may be repeated
Now you can choose any of 10 digits to be first, any of those same 10 digits to be next, etc. There are $\displaystyle 10^6$ such digits.

c) at least one 5 is included
If digits can also be repeated, which I assume since you say "at least one 5 is included", then you can do this by selecting a 5 first, then any of 10 digits for the remaining 5 digits: that is $\displaystyle 10^5$. Since that single 5 can be in any of the 6 places, the total number of such numbers is $\displaystyle 6(10^5)$.

d) the number is even,greater than or equal to 600,000, with repetition of digits allowed
The first digit must be 6, 7, 8, 9, four choices and the last digit must be one of 0, 2, 4, 6, 8, five choices. The other 4 digits can be any of the 10 digits. There are $\displaystyle 4(5)(10^5)= 2000000$ such numbers.

As usual Soroban got in ahead of me. Blast him!

4. Originally Posted by Soroban
Hello, william!

I will assume that the number may not begin with 0 (zero).

The first digit cannot be zero: 9 choices.

The second digit can be any of the other 9 other digits.
The third digit can be any of the other 8 digits.
The fourth digit can be any of the other 7 digits.
The fifth digit can be any of the other 6 digits.
The sixth digit can be any of the other 5 digits.

Answer: .$\displaystyle 9 \times 9 \times 8 \times 7 \times 6 \times 5 \:=\:130,\!080$

The first digit has 9 choices.
Each of the other five digits has 10 choices.

Answer: .$\displaystyle 9 \times 10^5 \:=\:900,\!000$

In part (b), we saw that there are $\displaystyle 900,\!000$ possible numbers.

Consider the numbers that have no 5's.

The first digit has 8 choices (not 0 and not 5).
Each of the other five digits has 9 choices.
Hence, there are: .$\displaystyle 8 \times 9^5 \:=\:472,392$ numbers with no 5's.

Answer: .$\displaystyle 900,\!000 - 472,\!392 \:=\:427,608$

The number is $\displaystyle \geq 600,\!000$
The first digit must be: 6, 7, 8, or 9 . . . 4 choices.

The number is even.
The last digit must be: 0, 2, 4, 6, 8 . . . 5 choices.

Each of the other four digits has 10 choices.

Answer: .$\displaystyle 4 \times 5 \times 10^4 \:=\:200,\!000$

thank you! for a) you mean 136,080, right?