How many numbers of $7$ digits can be formed with the digit $0,1,1,5,6,6,6$.
My attempt:
Seventh place, total number of possibility is $=\frac{6!}{2!\times 3!}=60$ ways.
Sixth place, total number of possibility is $=\frac{7!}{2!\times 3!}=420$ ways.
fifth place, total number of possibility is $=\frac{7!}{2!\times 3!}=420$ ways.
Fourth place, total number of possibility is $=\frac{7!}{2!\times 3!}=420$ ways.
Third place, total number of possibility is $=\frac{7!}{2!\times 3!}=420$ ways.
Second place, total number of possibility is $=\frac{7!}{2!\times 3!}=420$ ways.
First place, total number of possibility is $=\frac{7!}{2!\times 3!}=420$ ways.
Therefore,
Total possible numbers are $60\times 420\times 420\times 420\times 420\times 420\times 420 = 32934190464\times10^7$.
( Ohh, Seriously ! ).
Can you explain in formal/alternative way, please?
Best Answer
There are $7$ symbols, that can be arranged in $7!$ many ways.
However, the three $6$'s and $2$ $1$'s can be interchanged, so we have $\frac{7!}{3! 2!} = 420$ many such sequences. But I think $0$ at the start is forbidden , so we have to substract all sequences that start with $0$, of which there are $\frac{6!}{3! 2!} = 60$. That leaves 360 numbers.