#Combinatorics #Self doubt

Question

How many 3 digits number are there which are divisible by 3 and repetition of digits NOT allowed.?

Comments

You can make 3 sets based on the equivalence classes of remainders 0,1 and 2 for all the digits 0 to 9.

$A= \{0,3,6,9\}$

$B= \{1,4,7\}$

$C = \{2,5,8\}$

Now, if you take one number from each set and add those numbers then it will be divisible by $3.$ The reason is they are equivalence classes of remainders 0,1 and 2 and so, if you add them, they will be divisible by $3.$

Now, there are $4$ choices to select a number from set $A,$ similarly, $3$ choices to select a number from set $B$ and $3$ choices to select a number from set $C.$

But since, it is a 3-digit number, so zero can’t come at the hundredth place. So, if we select a number from set A, initially, there will 3 choices.

So, 3 digits number without repetition and divisible by 3 are:

$1) ABC = 3 \times 3 \times 3$

$2) ACB = 3 \times 3 \times 3$

$3) BAC = 3 \times 4 \times 3$  

$4) BCA = 3 \times 3 \times 4$

$5) CAB = 3 \times 4 \times 3$

$6) CBA = 3 \times 3 \times 4$

Hence, answer is: $(2 \times 3^3) + (4 \times 4 \times 3^2)$

Edit:

As Jiren has mentioned that we have to look individual sets also. The reason is, the sets $B$ and $C$ are having elements with equivalence class $1$ and $2$ respectively and since these sets contain $3$ elements, so, $1+1+1=3$ and $2+2+2=6$ and 3 and 6 are divisible by 3. If these sets contain number of elements which are not multiple of 3 then we don't have to look for these sets.

So, from sets $B$ and $C$,  we can make $3!$ numbers according to the requirement of the question.

Now, since set $A$ is having $4$ elements and we don't have to make 0 at the hundredth place. So, first we can make 3 digit number from total 4 elements of set $A$ which may or may not have 0 at the hundredth place and then we subtract the 3 digit numbers which is having 0 at the hundredth place by fixing 0 at the hundredth place. So, total numbers will be $\binom{4}{3}3! \  - \  \binom{3}{2}2!$

Hence, Answer is  $(2 \times 3^3) + (4 \times 4 \times 3^2) + 3! + 3! +  \binom{4}{3}3! \  - \ \binom{3}{2}2!$

Credit: [JIren]

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@ankitgupta.1729 sir 

u are missing few cases  [ to be precise 30 cases ]

1. we can have 3 elements from the given set A itself like 0+6+9 = 15 

               so there will be C(4,3) * 3!  –  C(3,2) * 2! = 18 [ we should remove numbers starting from 0 ]

2. we can have all 3 elements from set B itself 1+4+7 = 12 so we have 3! ways i.e 6
3. we can have all 3 elements from set C itself 2+5+8 = 15 so we have 3! ways i.e 6

finally 18+6+6 = 30 So answer should be 198+30 = 228

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Yes. Thank you.

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wonderful explanation

Answer 1

Divide the digits into four sets:

S1 = $\{0\}$, S2 = $\{3,6,9\}$, S3 = $\{1,4,7\}$ and S4 = $\{2,5,8\}$

There can be a total of $2^4=16$ combinations of the following four sets. We have to choose the set combinations which result in digits summing up to a multiple of 3. We can take care of unique digits after that.

Case 1:  Digits belonging to $S3$ only: The sum of the digits is $12$. All digits are unique, so the number of possible orderings is $3! = 6$. 

Case 2:  Digits belonging to $S4$ only: The sum of the digits is $15$. All digits are unique, so the number of possible orderings is $3! = 6$.

Case 3: Digits belonging to $S1 \cup S2$ only: The possible sums are: $9$, $12$, and $18$. $0$ can’t be at the hundred’s place. The number of possible orderings is $3 * 3 * 2 = 18$.

Case 4: Digits belonging to $S1 \cup S3 \cup S4$ only: Here, we take exactly one digit from each of S1, S3, and S4. $0$ can be at either the ten’s or one’s place. So, the number of possible orderings is $2 * 6 * 3 = 36$.

Case 5: Digits belonging to $S2 \cup S3 \cup S4$ only: Again we take exactly one digit from each of S2, S3 and S4. The number of possible orderings is $9 * 6 * 3 = 162$.

 

Hence the count of three-digit numbers divisible by $3$ and without any repetition of digits is $228$.

Comments

The three digit natural numbers start from 100 and ends with 999.

The first three digit number which is divisible by 3 is 102.

The last three digit number which is divisible by 3 is 999.

The number of three digit numbers divisible by 3 are (999/3)​−(102/3)​+1=333−34+1=300.

 

What is wrong in this ?

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@kuchtohehi45 bro in the question they said not repeat the digits right 

in your collection 333 is allowed but it should not be counted 

you have answered to the question :  

How many 3 digits number are there which are divisible by 3 and repetition of digits are allowed.?

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Oh correct!

Areee can you please help how to deal with this type of mistake?