Permutations with restrictions and repitition

Question

Answer the following regarding three digit numbers if 234 is considered a 3 digit number but 034 is not:

a) How many odd three digit numbers are there if numbers can be repeated?

b) How many odd three digit numbers are there if no repeating numbers are allowed?

c) How many even three digit numbers are there if numbers can be repeated?

b) How many even three digit numbers are there if no repeating numbers are allowed?

Answer 1

Permutations with restrictions and repetition

Answer the following regarding three digit numbers if 234 is considered a 3 digit number but 034 is not:

a) How many odd three digit numbers are there if numbers can be repeated?

The answers for a and b are almost identical and fairly simple:

There are nine possible choices for the first digit, ten choices for the second, and five choices for the third digit, so there are 450 odd three-digit numbers--as calculated by the following:

9 x 10 x 5 = 450

b) How many odd three digit numbers are there if no repeating numbers are allowed?

Questions b and d get a little more involved on determining the third digit because zero can not be chosen for the first digit. That means there are five possible odd numbers to choose from for the first digit, but only four possible even numbers from which to choose.

Zero can be chosen for the second digit, but you cannot repeat the number that you chose for the first digit. Therefore, there are nine possible choices for the second digit.

Now, comes the tricky part:

If your first two digits were both even, you could choose any of the five odd numbers for the third digit. However, if your first two digits were both odd, you would only be allowed to choose between the three odd numbers that you had not chosen. If you had chose one of each for the first two digits, you would still have four odd numbers from which to choose for the third digit. There are four different combinations, so..let's make a table to show the possibilities:

| ------ Digit Types ---- | Poss. | --Type -- | Possible | Possible |_ Sub _|

_____________ | 1st digit | 2nd digit | Ways | 3rd digit |_ Ways _ | _Ways _ | Totals_|

-- Combo 1:_ | -- odd -- | --- odd --- | 5 x 4 | --- odd --- | ---- 3 ---- | 5 x 4 x 3 |__ 60 _|

-- Combo 2:_ | -- odd -- | ---even--- | 5 x 5 | --- odd --- | ---- 4 ---- | 5 x 5 x 4 | _100 _|

-- Combo 3:_ | -- even - | --- odd --- | 4 x 5 | --- odd --- | ---- 4 ---- | 4 x 5 x 4 |__ 80 _|

-- Combo 4:_ | -- even - | ---even--- | 4 x 4 | --- odd --- | ---- 5 ---- | 4 x 4 x 5 |__ 80 _|

_________________ Total odd three-digit numbers with no repeating digits: _320_<<< Answer

There are nine possible choices for the first digit and nine choices for the second. digit. and five choices for the third digit, so there are 450 even three-digit numbers--as calculated by the following:

9 x 10 x 5 = 450

c) How many even three digit numbers are there if numbers can be repeated?

There are nine possible choices for the first digit, ten choices for the second, and five choices for the third digit, so there are 450 even three-digit numbers--as calculated by the following:

9 x 10 x 5 = 450

d) How many even three digit numbers are there if no repeating numbers are allowed?

As discussed above, Questions b and d get a little more involved on determining the third digit because zero can not be chosen for the first digit. That means there are five possible odd numbers to choose from for the first digit, but only four possible even numbers from which to choose.

Zero can be chosen for the second digit, but you cannot repeat the number that you chose for the first digit. Therefore, there are nine possible choices for the second digit.

Now, comes the tricky part for this question:

If your first two digits were both odd, you could choose any of the five even numbers for the third digit. However, if your first two digits were both even, you would only be allowed to choose between the three even numbers that you had not chosen. If you had chose one of each for the first two digits, you would still have four even numbers from which to choose for the third digit. There are four different combinations, so..let's make a table to show the possibilities:

| ----- Digit Types ----- | Poss. | --Type --- | Possible | Possible |_ Sub _|

_____________ | 1st digit | 2nd digit | Ways | 3rd digit_|_ Ways _ |_ Ways _ | Totals_|

-- Combo 1: _| -- odd -- |_---odd---- |_5 x 4 | -- even -- | ---- 5 ---- | 5 x 4 x 5 | _100 _|

-- Combo 2: _| -- odd -- | -- even--- |_5 x 5 | -- even --- | ---- 4 ---- | 5 x 5 x 4 | _100 _|

-- Combo 3: _| --even - | ----odd---- |_4 x 5 | -- even --- | ---- 4 ---- | 4 x 5 x 4 |__ 80 _|

-- Combo 4: _| --even - | -- even--- |_4 x 4 | -- even --- | ---- 3 ---- | 4 x 4 x 3 |__ 48 _|

________________ Total even three-digit numbers with no repeating digits: _348_<<< Answer