# Thread: Number of 6 digit numbers

1. ## Number of 6 digit numbers

find the number of 6 digit numbers that can be made with digits 1,2,3,4 if all digits are to appear in the number atleast once

2. There are six spots to consider, how many combinations for each?

All numbers need to appear at least once and for the last two spots they can be the same or different. The answer will be of the form.

$4\times 3\times 2\times 1\times 4 \times 4+4\times 3\times 2\times 1\times 4 \times 3$

Do you follow?

3. Originally Posted by prasum
find the number of 6 digit numbers that can be made with digits 1,2,3,4 if all digits are to appear in the number atleast once
You need to count the number of surjections from a set of six to a set of four.
$\text{Surj}(n,k) = \sum\limits_{j=0}^{k} {( - 1)^j \binom{k}{j} \left( {k - j} \right)^n }$.

4. Hello, prasum!

Find the number of 6-digit numbers that can be made with digits {1, 2, 3, 4}
if all digits are to appear in the number at least once.

Let the four digits be: $\{a,\,b,\,c,\,d\}$

Select one of each digit: . $a\,b\,c\,d\,\_\,\_$

There are two cases to consider.

[1] The other two digits are the same: . $4$ choices.
. . .The number is of the form: $a\,a\,a\,b\,c\,d$
. . .The digits can be permuted in: ${6\choose3} = 20$ ways.
. . .Hence, there are: . $4 \times 20 \:=\:80$ such numbers.

[2] The other two digits are different: . ${4\choose2} \,=\, 6$ choices.
. . .The number is of the form: $a\,a\,b\,b\,c\,d$
. . .The digits can be permuted in ${6\choose2,2} = 180$ ways.
. . .Hence, there are: . $6 \times 180 \:=\:1080$ such numbers.

The answer is: . $80 + 1080 \:=\:1060$ numbers.

5. Originally Posted by Soroban
Hello, prasum!

Let the four digits be: $\{a,\,b,\,c,\,d\}$

Select one of each digit: . $a\,b\,c\,d\,\_\,\_$

There are two cases to consider.

[1] The other two digits are the same: . $4$ choices.
. . .The number is of the form: $a\,a\,a\,b\,c\,d$
. . .The digits can be permuted in: ${6\choose3} = 20$ ways.
. . .Hence, there are: . $4 \times 20 \:=\:80$ such numbers.

[2] The other two digits are different: . ${4\choose2} \,=\, 6$ choices.
. . .The number is of the form: $a\,a\,b\,b\,c\,d$
. . .The digits can be permuted in ${6\choose2,2} = 180$ ways.
. . .Hence, there are: . $6 \times 180 \:=\:1080$ such numbers.

The answer is: . $80 + 1080 \:=\:1060$ numbers.

thanks

6. I might point out that the correct answer is 1560.

7. Originally Posted by Soroban

. . .The digits can be permuted in: ${6\choose3} = 20$ ways.

it is 120, not 20 (you calculated combinations rather than permutations)