The number of $n$ digit numbers which consists of the digits $1$ and $2$ only if each digit is to be least once, is equal to $510$ then find the value of $n$ .

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Solution

Basically we have n blank spaces in which we have to put $1$ and $2$

Total such occurrence $= 2 n$
But we have to subtract the cases where there is no occurrence of $1$ and no occurrence of $2$

Hence, Number with at least one occurrence

We have $2^{n} - 2 = 510$ ;
$\Rightarrow 2^{n} = 512$
$\Rightarrow n = 9$ .

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The number of n digit numbers which consists of the digits 1 and 2 only if each digit is to be least once, is equal to 510 then find the value of n.

Basically we have n blank spaces in which we have to put 1 and 2Total such occurrence=2nBut we have to subtract the cases where there is no occurrence of 1 and no occurrence of 2Hence, Number with at least one occurrenceWe have 2n−2=510;⇒2n=512⇒n=9.

The number of $n$ digit numbers which consists of the digits $1$ and $2$ only if each digit is to be least once, is equal to $510$ then find the value of $n$ .