Combinatorics..#2

Probability Level 3

How many 4 digit numbers can be formed with the digits $1 , 3 , 3 , 0.$

Each digit can only be used once.

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The answer is 9.

3 solutions

Sonali Srivastava
Jul 18, 2014

1st 2nd 3rd 4th positions on 1st position we can't place 0 so any one of the remaining digits can be placed(1,3,3) ...on 2nd position one digit from remaining digits can be placed ..on 3rd one of the remaining 2 n on 4th the last one...so no. of possibilities on each position is multiplied to give 3 3 2*1=18 but digit 3 occurs twice hence one nos. are repeated so 18 is divided b y the no. of repeating digits i.e. 2 18/2=9 ans.

Harshvardhan Mehta
Jul 17, 2014

The required number of number = $\frac{4!}{2!} - \frac{3!}{2!} = 9$

Question is not completed ,one digit can be used multiple time

prabhat sharma - 6 years, 11 months ago

sry i dint get you..

Harshvardhan Mehta - 6 years, 10 months ago

you got it and edit the question by putting last line on it

prabhat sharma - 6 years, 10 months ago

@Prabhat Sharma – ohhh yeah understood...

Harshvardhan Mehta - 6 years, 10 months ago

But how can you make a 4 digit number according to the rules given??

Anuj Shikarkhane - 6 years, 10 months ago

i am sorry but i dint get you

Harshvardhan Mehta - 6 years, 10 months ago

Can you explain your solution in more detail? How are those numbers calculated? Thanks

Calvin Lin Staff - 6 years, 5 months ago

well its like $\frac{4!}{2!}$ because of the 4 digits and 2! because there are 2 3's so they are the total number of 4 digit numbers formed by these digits , and we subtract it by $\frac{3!}{2!}$ so as to remove the numbers which start with 0 . and 2! as before.

Harshvardhan Mehta - 6 years, 4 months ago

Mohit Khanwale
Feb 7, 2015

For thousands place we can't place 0 so we have 3 choices , next for hundreds place we are left with again 3 choices , for tens place we have 2 choices and for units 1 choi ce,so we have 3 ×3×2× 1= 18 . But we have divide by 2! Because of Two 3s . So we finally he 18/2 = 9.…..…...I hope you liked the solution.😀😁

Combinatorics..#2

The answer is 9.

3 solutions

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