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Re: Permutation combination
by Ankit Kaushik - Thursday, 19 November 2015, 02:01 AM
 

For single digit,there will be nine numbers having a single digit. For two digits, there will be two types of numbers : abbb & aabb. First let's choose two digits,it can be done in C(10,2) ways.

Now for first type abbb, arrangement of digits can be done in { 2xfac(4)/fac(3)}.Here, we have first shuffled both digits in abbb and then divided by fac(3).Also,the two digits we chose can alternately take the variable a & b.

Now,for the type aabb, we will first shuffle the digits i.e fac(4) and the divide them by fac(2) two times.(cos there are 2 a's & 2 b's).Guess why ,we are not here multiplying by 2.

So the answer will be :

9 + C(10,2) x{ [(2xfac(4)/fac(3))] + [fac(4)/(fac(2)xfac(2))] }