Permutation-Combination
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Q.

How many five digit positive integers that are divisible by 3 can be formed using the digits $0, 1, 2, 3, 4$ and $5$, without any of the digits getting repeating?

 A.

15

 B.

96

 C.

216

 D.

120

 E.

625

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Solution:
Option(C) is correct

Test of divisibility for 3

The sum of the digits of any number that is divisible by '3' is divisible by 3. 

For instance, take the number 54372.

Sum of its digits is,

$=5 + 4 + 3 + 7 + 2$

$= 21$

As 21 is divisible by '3', 54372 is also divisible by 3.

There are six digits viz., $0, 1, 2, 3, 4$ and $5$. To form 5-digit numbers we need exactly 5 digits. So we should not be using one of the digits. 

The sum of all the six digits $0, 1, 2, 3, 4$ and $5$ is $15$. We know that any number is divisible by 3 if and only if the sum of its digits is divisible by '3'. 

Combining the two criteria that we use only 5 of the 6 digits and pick them in such a way that the sum is divisible by 3, we should not use either '0' or '3' while forming the five digit numbers. 

Case 1

If we do not use '0', then the remaining 5 digits can be arranged in:

$5!$ ways$ = 120$ numbers.

Case 2

If we do not use '3', then the arrangements should take into account that '0' cannot be the first digit as a 5-digit number will not start with '0'.

The first digit from the left can be any of the 4 digits $1, 2, 4$ or $5$ 

Then the remaining 4 digits including '0' can be arranged in the other 4 places in $4!$ ways.

So, there will be $4×4!$ numbers $= 4×24 = 96$ numbers.

Combining Case 1 and Case 2, there are a total of $120 + 96 = \textbf{216}$, 5 digit numbers divisible by '3' that can be formed using the digits 0 to 5.


(6) Comment(s)


Johny
 ()

Case 1 when we do not include 0

we get 5! = 120 ways

Case 2-> Suppose i want to include 0 but not at the first position because

five digits number includes 0 but not at first position........

so let 0 is at last position remaining digits are 1,2,3,4 and 5

if i make a number from this say 54320 but this is not divisible by 3

in order to make it divisible by 3 i have to exclude 3 and add 1 so the

number becomes 54120 which is divisible by 3

5412 can be arranged in 4! ways

similarly if 0 is at second position from right we get another 4! ways

thus for four positions of 0 we get 4*4! (4!+4!+4!+4!)ways

so total ways =120+(4*4!)



Cindy
 ()

Why should we choose to not use either 0 or 3 ?



Mahi
 ()

hey man wht procedure tht was,u have to consider it's divisibility rule also yaar



Chaitana
 ()

sorry, I got the answer.



Chaitana
 ()

I have a question.

Solution I get is 600:

0 1 2 3 4 5

first we select only from 5 digits because we cannot have zero as the first digit for five digit number.

first digit -> 5 ways

Repetition is not allowed

one of the digits from 1 2 3 4 5 is already selected. So, we have only 5 digits left because we can now use 0 as well

second digit - 5

third digit - 4

fourth digit - 3

fifth digit - 2

In total it would be: $5*5*4*3*2 = 600$.

What is wrong with this procedure?

Why do I get a different answer.


Guglani
 ()

Here you only created a five digit no. with this explanation, but its not divisible by 3.