COMBINATORICS

To be divisible by 5 and 2 a whole (natural) number should be factored to both 2 and 5 (both of them are primes, one cannot be divided by other)- hence we have 2*5 OR 10. Basically all our numbers should have ending of 0 (zero). Since we may not have repetitions here - use permutations starting with the sixth digit in the first place --> 6 (6 digits can be here excluding 0) * NOW switch to the last digit (as 0 cannot be repeated) ~~ 6*X*X*...*1 (for 0), then switch to the second digit again ...
6*5*4*3*2*1=6! OR 720 ways (whole numbers) in total

gmat7202011 wrote:1. How many even natural numbers divisible by 5 can be formed with digits 0,1,2,3,4,5 and 6 ( no repetitions)

Thank You,

gmat7202011 wrote:1. How many even natural numbers divisible by 5 can be formed with digits 0,1,2,3,4,5 and 6 ( no repetitions)

Thank You,

one digit even natural no. that are divisible by 5 = 0;

two digit even natural no. that are divisible by 5= 6;(6*1) as the unit digit should be zero and the the tens place can be filled with any of the six digits.

three digit even natural no. that are divisible by 5 = 30; ( 6*5*1) as the unit digit should be zero therefore no. of ways of filling it is 1, and the tens and hundred place can be filled with any of the remaining 6 numbers in 6*5 ways =30;

four digit even natural no. that are divisible by 5 = 120 (6*5*4*1);

five digit even natural no. that are divisible by 5 = 360(6*5*4*3*1);

six digit even natural no. that are divisible by 5 = 720(6*5*4*3*2*1);

seven digit even natural no. that are divisible by 5 = 720 (6*5*4*3*2*1*1);

hence total no. of even natural no. that are divisible by 5 are 720+720+360+120+30+6 = 1956

Let me get back after checking the answer today evening when i go home

Thank You,
Yogesh

We need to find the sum of single digit, two digit ,.....and seven digit numbers

For it to be even and divisible by 5, it should be divisible by both 2 and 5, and should end with only 0. So the last digit is fixed to 0. There is only way to pick the units digit as per the given condition.

i) Seven Digit Numbers = 6C6*6! = 6! (last number is 1, so it does not make an effect)
ii) Six Digit Numbers = 6C5 * 5! = Out of the six numbers(1,2,3,4,5,6) you have you pick 5 numbers and the last digit is zero. The no of ways is nCr * r! (to pick r numbers from n and then arrange them)
Similarly
iii) Five Digit Numbers = 6C4*4!
iv) Four Digit Numbers = 6C3*3!
v) Three Digit Numbers = 6C2*2!
vi) Two Digit Number = 6C1*1!

So the to total No of ways are :
6!+6C5*5!+6C4*4!+6C3*3!+6C2*2!+6
= 720+720+360+120+30+6 = 1956

Does anybody have a shorter method?

the solutions by other fellows helped me correct my solution as I have counted only 6-digit numbers, 5-digit and the remaining numbers must be counted too. Total makes 1956 numbers.

Night reader wrote:To be divisible by 5 and 2 a whole (natural) number should be factored to both 2 and 5 (both of them are primes, one cannot be divided by other)- hence we have 2*5 OR 10. Basically all our numbers should have ending of 0 (zero). Since we may not have repetitions here - use permutations starting with the sixth digit in the first place --> 6 (6 digits can be here excluding 0) * NOW switch to the last digit (as 0 cannot be repeated) ~~ 6*X*X*...*1 (for 0), then switch to the second digit again ...
6*5*4*3*2*1=6! OR 720 ways (whole numbers) in total

gmat7202011 wrote:1. How many even natural numbers divisible by 5 can be formed with digits 0,1,2,3,4,5 and 6 ( no repetitions)

Thank You,