IIT JEE , AIEEE Forum - Mathematics , Algebra - Maths Olympiad ---> PnC

sandymaurya

Since every no. (from 1 to 5) can be repeat so there are 10 no which can be used.
i.e. 1,1,2,2,3,3,4,4,5,5

but we need only a no. containg 6 digit out of 10 digits in which three are repeated.
Or we can say the total no. of combinations which can be formed are from five digits and we have to choose only three digits, because when we choose a digit say 1 than by condition we should have to choose another 1.
In sort you can say we have to choose in pairs.
Thus total no. of combinations possible are ⁵C_{3 = 10.

Now for each combination we have maximum arrangements with repetitions are 6!.} So we have a formula for repeated permutations.

(no. of possible outcomes with repetitions) / p_{1! *} p_{2! *} ... _* p_n!

where p_1, p_2, and p_n are repeatitions

thus for each combination

P₁ = 6!/(2!_*2!_*2!)=630

and for 10 combinations

P = 10 x P₁ = 10 x 6! / ( 2! x 2! x 2! ) = 100

IF U FIND THIS USEFUL FOR U AND ALSO IF IT IS CORRECT THAN RATE IT

Ask & Discuss Questions with Community & Experts