# Thread: Probability using permutation and combination

1. ## Probability using permutation and combination

Question : Four-figure numbers are to be formed from the digits 4,5,6,7,8,9 .How many different four-figure numbers can be formed if there is at least one repeated digit, but no digit appears more than twice in the number.
trying a lot but not getting the answer.plz help

2. ## Re: Probability using permutation and combination

You can do this in two parts. Two pairs or one pair and two different.

If you have two pairs you have 6C2 choices for what the digits are. These can be arranged 4!/(2!2!) ways.

If there is only one pair you are choosing three digits with three choices for which is doubled. That's 6C3x3. These can be arranged 4!/2! ways.

Final answer :$\displaystyle \frac{\binom{6}{2} \times 4!}{2! 2!}+\frac{\binom{6}{3}\times 3 \times 4! }{2!}=90+720=810$

3. ## Re: Probability using permutation and combination

when there's only one pair...
why should we multiply by 3 ?

4. ## Re: Probability using permutation and combination

For 1 repeated digit... i did this.
6P1 x 6P1 x 5P2 = 720

For 2 repeated digits,
6P1 x 6P1 x 5P1 x 5P1 = 900

the total doesnt equal 810.
I think my mistake lies in the fourth line, concerning the 2 repeated digits.
Could you correct me ?

5. ## Re: Probability using permutation and combination

Originally Posted by nks2427
For 1 repeated digit... i did this.
6P1 x 6P1 x 5P2 = 720
For 2 repeated digits,
6P1 x 6P1 x 5P1 x 5P1 = 900
the total doesnt equal 810.
I think my mistake lies in the fourth line, concerning the 2 repeated digits.
Could you correct me ?
It does not equal to 810 because the reasoning is incorrect.
Here is one way $\displaystyle 6^6-\frac{6!}{2!}-\binom{6}{1}\binom{5}{1}\frac{4!}{3!}-6=810$
What I think reply #2 means is $\displaystyle \binom{6}{1}\binom{5}{2}\frac{4!}{2!}+\binom{6}{2} \frac{4!}{(2!)^2}=810.$