Permutation Problems #2

• Oct 23rd 2009, 01:03 PM
saberteeth
Permutation Problems #2
(Hi),

How do you solve these?

1) A number of 4 different digits is formed using the digits 1,2,3,4,5,6,7. How many numbers can be formed that are greater than 3400?

2) How many numbers of different digits lying between 100 and 1000 can be formed with the digits 2,3,4,0,8,9?
• Oct 23rd 2009, 02:08 PM
Soroban
Hello, saberteeth!

We have to "talk" our way through these . . .

Quote:

1) A number of 4 different digits is formed using the digits {1,2,3,4,5,6,7}.
How many numbers can be formed that are greater than 3400?

There are two cases to be considered:
. . [1] The first digit is 3.
. . [2] The first digit is 4 or more.

[1] The 1st digit is 3: .$\displaystyle 3\:\_\:\_\:\_$
. . .The 2nd digit must be {4,5,6,7} . . . 4 choices.
. . .The 3rd digit has 5 choices.
. . .The 4th digit has 4 choices.
There are: .$\displaystyle 4\cdot5\cdot4 \:=\:{\color{blue}80}$ numbers greater than 3400 which begin with 3

[2] The 1st digit is 4 or more.
. . .The 1st digit must be {4,5,6,7} . . . 4 choices.
. . .The 2nd digit has 6 choices.
. . .The 3rd digit has 5 choices.
. . .The 4th digit has 4 choices.
There are: .$\displaystyle 4\cdot6\cdot5\cdot4 \:=\:{\color{blue}480}$ other numbers greater than 3400.

Therefore, there are: .$\displaystyle 80 + 480 \:=\:\boxed{560}$ numbers greater than 3400.

Quote:

2) How many numbers of different digits lying between 100 and 1000
can be formed with the digits {0,2,3,4,8,9}?

These are all 3-digit numbers.

The first digit can be: {2,3,4,8,9} . . . 5 choices.
The second digit has 5 choices.
The third digit has 4 choices.

Therefore, there are: .$\displaystyle 5\cdot5\cdot4 \:=\:\boxed{100}$ such numbers.

• Oct 23rd 2009, 09:12 PM
saberteeth
Cool, thanks :) I got it.