Permutations and Combinations? 3 digit number... 1) Make a 3 digit even number without repeated digits, using 0, 4, 5 , 6, 7. Also the first digit cannot be 0.
2)Arrange 12 books in a line, 4 of which are english, 3 of which are science, and 5 calculus, so that all books of same subject are adjacent.
 A: Let's take your first question:
case01:Repetition is not allowed
You can see that in a three digit number there are three place ie. One ten and hundred
Now hundred's place can have 4 numbers 4,5,6 or 7.so 4 ways
Now ten place again 4 numbers can be accommodate 0 , and from other 3 remaining because repetition is not allowed. so 4 ways 
Now ones place 3 numbers can be accommodate so 3 ways 
Now total ways would be 4x4x3=48 numbers are possible if repetition is not allowed.
case02: Repetition is allowed 
for hundreds place 4 numbers can accommodate , for tens place 5 numbers can accommodate and for ones place once again 5 numbers can accommodate. (here in hundreds place we can't use 0 so we have 4 numbers in all other each of the number is used)
so total ways when repetition us allowed is 4x5x5=100.
Let's look at your second question 
Let's assume that you first arranged English 4 book 
ways to arrange them is 4! = 24
then you put science 3 books
ways to arrange it would be 3!=6
then you put calculus 5 books
ways to arrange it would be 5!=120
Now you have three groups science calculus and English hence ways to arrange these three groups is 3!=6.
total ways =120x6x24x6=103680.
A: We can divide into $2$ cases, Case $1$, where the first (leftmost) digit is even, and Case $2$, where the first digit is odd. 
Case 1: There are $2$ possibilities for the first digit, since $0$ is forbidden.  For each of these, there are $2$ possibilities for the last digit, since it must be even.  And now there are $3$ digits left, any one of which can be used for the middle digit, giving $(2)(2)(3)$ numbers.
Case 2: There are $2$ choices for the first digit. For each of these, there are $3$ choices for the last digit, and then $3$ for the middle digit, giving  $(2)(3)(3)$ numbers.
Add. 
A: On case 1: 
First we will calculate how many numbers there are if the condition that the number is even is left out. For the first digit there are $4$ choices, for the second there are $4$ choices and for the third there are $3$ choices. So this results in $4\times 4\times 3=48$ possibilities. 
If such a number is odd then it ends on $5$ or $7$. 
How many numbers of the $48$ numbers end on $5$? Now for the first digit there are $3$ choices, for the second there are $3$ choices, showing that $3\times 3=9$ end on a $5$. Likewise we find that $9$ end on a $7$ and subtraction gives $48-9-9=30$ numbers that are even.
