Probability using permutation and combination

• Sep 10th 2012, 10:15 PM
nks2427
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
• Sep 11th 2012, 01:31 AM
a tutor
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 : $\frac{\binom{6}{2} \times 4!}{2! 2!}+\frac{\binom{6}{3}\times 3 \times 4! }{2!}=90+720=810$
• Sep 11th 2012, 09:33 AM
nks2427
Re: Probability using permutation and combination
when there's only one pair...
why should we multiply by 3 ? :(
• Sep 11th 2012, 09:42 PM
nks2427
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 ?
• Sep 12th 2012, 06:23 AM
Plato
Re: Probability using permutation and combination
Quote:

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 $6^6-\frac{6!}{2!}-\binom{6}{1}\binom{5}{1}\frac{4!}{3!}-6=810$
What I think reply #2 means is $\binom{6}{1}\binom{5}{2}\frac{4!}{2!}+\binom{6}{2} \frac{4!}{(2!)^2}=810.$