1. ## combinatorics

How many strings of five decimal digits
a. begin with an odd digit?
my answer) $5.10^3$
b. do not contain the same digit twice?
my answer) $10.9.8.7.6$
c. have exactly three digits that are 9s?
my answer) $90$

that's 5.2=10 places for remaining 9 digits
and 9 choices for each place = 10x9 = 90, but,
can't it be like this....
exactly 3 same digits means $9^3.10^2$?

d. contain exactly three distinct digits?

2. In a) $5\cdot 10^4$

In b) can the first digit be zero? Such as 06453.
If not then it should $9 \cdot 9 \cdot 8 \cdot 7 \cdot 6$

3. Originally Posted by Plato
In a) $5\cdot 10^4$

In b) can the first digit be zero? Such as 06453.
If not then it should $9 \cdot 9 \cdot 8 \cdot 7 \cdot 6$
Can you answer the confusion that I am having with c)?

Can the first digit be zero as in 06236??

5. Originally Posted by Plato
Can the first digit be zero as in 06236??
It hasn't been explicitly mentioned, so I guess 0 counts too.

6. Assume that the number can have the begin with a zero.

c. have exactly three digits that are 9s?
${5 \choose 3} \cdot 9^2$.

7. Originally Posted by Plato
Assume that the number can have the begin with a zero.

c. have exactly three digits that are 9s?
${5 \choose 3} \cdot 9^2$.
What if it cannot begin with a zero? What will it be then?

Also can you answer the question c) without using "Combination" i.e. by only using "Product Rule"?

8. ${4 \choose 2} \cdot 9^2 + 8 \cdot {4 \choose 3} \cdot 9$ if the first is cannot be zero.

9. Hello, robocop_911!

Since it says "strings" rather than "five-digit numbers",
. . I assume they can begin with zero.

How many strings of five decimal digits

c. have exactly three digits that are 9's?

There are ${5\choose3}$ ways to place the 9's.
The other two places can be filled in $9^2$ ways.

My answer: . $10 \times 81 \:=\:8,100$

d. contain exactly three distinct digits?
This takes a bit more work . . .

The string is of the form: . $(1)\;aaabc\,\text{ or }\,(2)\;aabbc$

Form (1): . $aaabc$

There are ${10\choose3}$ choices for the three digits.
Then we decide which digit is the "triple": 3 choices.
And the five digits can be arranged in: . $\frac{5!}{3!}$ ways.

. . There are: . $120 \times 3 \times 20 \:=\:7,200$ strings of the form $aaabc.$

Form (2): . $aabbc$

There are ${10\choose3}$ choices for the three digits.
We we decide which two digits are the "pairs": 3 choices.
And the five digits can be arranged in: . $\frac{5!}{
2!2!}$
ways.

. . There are: . $120 \times 3 \times 30 \:=\:10,800$ strings of the form $aabbc.$

Therefore, there are: . $7,200 +10,800\:=\:{\color{blue}18,000}$ strings with exactly three distinct digits.

10. Originally Posted by Soroban

Since it says "strings" rather than "five-digit numbers",
. . I assume they can begin with zero.

There are ${5\choose3}$ ways to place the 9's.
The other two places can be filled in $9^2$ ways.

My answer: . $10 \times 81 \:=\:8,100$
Shouldn't it be P(5,3) since C(5,3) includes $9_1 9_2 9_3$

11. May I point out to you that Soroban's answer is exactly the same that I gave you. Once again, you are showing a real weakness with respect to definitions.