Solveeit Logo

Question

Question: The number of n digit numbers which consists of the digits 1 & 2 only if each digit is to be used at...

The number of n digit numbers which consists of the digits 1 & 2 only if each digit is to be used at least once, is equal to 510 then n is equal to ______.
(a). 7
(b). 8
(c). 9
(d). 10

Explanation

Solution

Hint: We will be using the concept of permutation and combination to solve the problem we will be using the multiplication principle to find the total possible number of cases.

Complete step-by-step solution -
Now, we have been given that a n-digit number which consists of digits 1 and 2 only if each digit is to be used at least once is 510.
Now, we will first find the total number n-digits possible with 1 and 2. For this we will use the multiplication principle in which we multiply all the cases for each digit. Here we have n digits so,


In any of the box we can have either 1 or 2, so we have 2 cases each box for all n boxes. Therefore, total n-digits number =2n={{2}^{n}} .
Now, we have also been given that 1 and 2 have to use at least once. So, we will subtract 2 cases when either all digits are 1 or all digits are 2, from total where all n digits are either 1 or 2.
Therefore, we have 2n2=510{{2}^{n}}-2=510
2n=510+2{{2}^{n}}=510+2
2n=512{{2}^{n}}=512 .
Now, we know that 512=29512={{2}^{9}}
2n=29{{2}^{n}}={{2}^{9}} .
On comparing powers we have n=9.
Hence the value of n is 9.

Note: To solve these types of questions we must know basic technique of permutation and combination like multiplication principle, addition principle to solve the problem quickly and easily.