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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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Is this a homework assignment? If not, can you post some of your own progress on the problem? –  Asaf Karagila Apr 21 '11 at 15:49
    
I really tried but i am unable to solve full...any way i will do some calculation Now..the i will post –  prem shekhar Apr 21 '11 at 16:01

2 Answers 2

up vote 1 down vote accepted

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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First digit cant be Zero –  prem shekhar Apr 21 '11 at 16:21
    
The main hardship for me is summing up the digits.I have tried this question making it short lik suppose only 1,2,3,4 are used and using this how many 3 digit numbers can be made and then what will be the sum..lik this –  prem shekhar Apr 21 '11 at 16:24
    
@prem: In this example, you have 4 choices for the first digit, 3 for the second, and 2 for the third, so there are 4*3*2=24 numbers. Of the 24, six will have 4 in the hundreds place, six will have 3, and so on. So the sum of the hundreds places is 6(1+2+3+4)100=6*10*100=6000. The sum of all the places will be 6(1+2+3+4)(100+10+1)=6660 –  Ross Millikan Apr 21 '11 at 16:40
    
@Ross Ya ya But here there is no concern about divisiblity is given....In my original question numbers must be divided by 5 and 9.............I have gone thru what you are saying.....THX –  prem shekhar Apr 21 '11 at 16:58
    
Try out how can we generalize the divisiblity like you have done it for summation –  prem shekhar Apr 21 '11 at 16:59

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.

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