Permutations and Combinations? 3 digit number...

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.