# Math Help - Simple permutation problems

1. ## Simple permutation problems

1.Find the numbers greater than 23000 that can be formed from the digits 1,2,3,4,5,6 without repeating any digit.

2.how many 6 digit numbers can be formed without repeatig any digit from the digits 0,1,2,3,4,5?
in how many of them will 0 be at tens place?

3.Find the number of 5 digit numbers that can be formed from the digits 1,2,4,6,8 (when no digit is repeated) but

i) the digits 2 and 8 are next to each other
ii) the digits 2 and 8 are not next to each other

Help will be appriciated.

2. ## Re: Simple permutation problems

The first one: the number has to be greater than 23000. So the number at ten thousand place can be occupied by any one of the numbers, 2,3,4,5,6 That is in 5 ways.
number at thousand place can be occupied by any one of the digits 3,4,5,6, that is in 4 ways The number at hundred place can be any one of the remaining 4 digits and tens place can be filled by any one of the remaining 3 digits and units place by remaining 2 numbers.
So we have total numbers greater than 23000 is given by 5x4x4x3x2

3. ## Re: Simple permutation problems

Hello, hacker804!

1.Find the numbers greater than 23,000 that can be formed
from the digits {1, 2, 3, 4, 5, 6} without repeating any digit.

There are a number of cases to consider.

Five-digit numbers

[1] The number begins with "23": . $2\;3\;\_\;\_\;\_\;\_$
. . .The other four digits can be arranged in $4! = 24$ ways.

[2] The number begins with "2": . $2\;\_\;\_\;\_\;\_\;\_$
. . .The second digit can be {4, 5, 6} . . . 3 choices.
. . .The other four digits can be arranged in $4! = 24$ ways.
. . .Hence, there are:. $3\cdot24 \:=\:72$ ways.

[3] The first digit is not "2": 5 choices.
. . .The other five digits can be arrnged in $5! = 120$ ways.
. . .Hence, there are:. $5\cdot5! \:=\:600$ ways.

Hence, there are:. $24 + 72 + 600 \:=\:696$ five-digit numbers.

Six-digit numbers

There are:. $6! = 720$ six-digit numbers.

Therefore, there are:. $696 + 720 \:=\:1416$ such numbers.