Permutation and Combination Problems

Question

Hey folks I'm having some issues with permutation and combination problems.

1) Make a 3 digit even number without repeated digits, using 0, 4, 5 , 6, 7. Also the first digit cannot be 0.

I figured it must be 4 X 4 x 3 = 48 because the first digit can be 4 digits. The second digit can also be 4 because you can use 0 but not the digit you already used and the third can be 3 because you can't use 5 or 7.

2)Arrange 12 blocks in a line, 4 of which are green, 3 of which are blue, and 5 red, so that all blocks are adjacent. No clue what this even means...

3)form a 5 digit number using each digit 1 through 5 so that 4 & 5 are not adjacent

4)Form a 5 digit without zeroes, with exactly four distinct digits (that is, one digit should be repeated).

Answer 1

Hint on 1):

• How many numbers can be made if it is not demanded that the number is even?

• How many of these numbers end up with a $5$? And how many with a $7$?

Answer 2

3) There are 5! way to permute the numbers. For each of these 120 permutations, there are 4 pairs of adjacent numbers. There are $\binom{5}{2}=10$ types of such pairs. That is for each permutation chances that 4 and 5 are adjacent are 4/10, hence the result becomes $5!*(1-4/10)$.

4) Firstly choose 4 numbers. There are $\binom{9}{4}$ ways.
Now choose one of the 4 numbers to double. There are $\binom{4}{1}$.
Now permute the numbers. There are $\binom{5}{2,1,1,1}=\frac{5!}{2!\cdot1!\cdot1!\cdot1!}$
The result is the product of these 3 numbers.