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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?

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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.

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There are 6 possible positions for the 0 (not the first). For each of these there are then 6 possible locations for the 5, so a total of 36 so far. There are now ${5\choose2}=10$ possible choices for the positions of the two 1s. The number is then completely determined, because the three 6s must go into the remaining positions.

So in total there are 360 possible numbers that can be formed.

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  • $\begingroup$ Thanks for nice explanation. $\endgroup$ – 1 0 May 29 '16 at 10:12
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Umm your answer is greater than the total possible 7 digit numbers even though you are not using all the possibilities ....

I think the correct way would be to use exclusion i.e all numbers that you can generate using those numbers - all possible numbers you generate that begin with 0. Therefore, numbers that can be generated using 0,1,1,5,6,6,6 without restriction are:

$$\frac{7!}{2!\times3!}=420$$

Now consider the numbers that can be generated from these numbers but start with 0. So first place is fixed and remaining 6 places we can decide

$$\frac{6!}{2!\times3!}=60$$

Removing these many numbers from our original set of numbers we arrive at 420-60=360 numbers which can be generated.

Please tell me if the answer is inccorrect. Hope I have helped !

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  • $\begingroup$ Thanks for nice explanation. $\endgroup$ – 1 0 May 29 '16 at 10:12

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