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Using the digits $0,1,2,3,4,5,6,7,8,9$, If five digit numbers is made without the repetition:

$1.$ How many numbers can be made?

$2.$ sum of all the even numbers?

$3.$ sum of all the odd numbers?

$4.$ How many numbers having unit digit as $5$?

$5.$ Sum of all the numbers which are divisible by $5$ and $9$ simultaneously?

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    I really tried but i am unable to solve full...any way i will do some calculation $N$ow..the i will post2011-04-21

2 Answers 2

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Hints: 1)How many choices for the first digit (is $0$ allowed?)? How many for the second? 2 and 3)If you add up all the numbers, each digit appears the same number of times in each column (assuming a leading $0$ is allowed). How many is that? How does that help? Now can you sort them somehow between evens and odds? 4)Do this after 1, the same way, starting from the units digit 5)What does it take for a number to be divisible by $5$? By $9$ How does that restrict your choices. This is by far the hardest.

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    For evens, you need the ones digit to be even, so it has to be 2 or 4. So there are 12 numbers that are even, six ending with 2 (and having 1,3,4 twi$c$e in each other place) and six ending with 4 (and having 1,2,3 twice in each other place). So the sum of the numbers is 2(1+3+4)(100+10)+2(1+2+3)(100+10)+6(2+4). Note how the evens become less common in the other places because you use them up in the ones place.2011-04-21
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The phenomenon you are looking for is sometimes loosely called the Multiplication Principle. Suppose you have $n$ different operations, and for clarity and ease let's call these operations $A_1, A_2, \dots , A_n$. For example, these operations could be picking out an outfit, picking a card from a deck, selecting letters or numbers for a license plate, etc. Furthermore, let's say there are $k_1$ ways to accomplish the operation $A_1$, $k_2$ ways to accomplish operation $A_2$, and so on. Then, the number of ways to accomplish all the operations $A_1, A_2, \dots , A_n$ is:

$ k_1 \times k_2 \times \dots \times k_n. $

For example, if you have $4$ shirts, $10$ pairs of pants, and $7$ pairs of shoes, the number of outfits you can make is

$ 4 \times 10 \times 7 = 280. $

If you have a hockey team with $24$ players, the number of ways to select a captain and assistant captain are

$ 24 \times 23 $

since there are $24$ players to be selected for the captaincy and $23$ players (everyone but the one already selected to be captain) for the assistant captaincy.

In light of this "multiplication principle" and Ross's comments, you should be able to write down the answers for each question. Only after you have tried these problems (and shared your attempts), ask for help if necessary.