Question

Total number of 6-digit number in which only and all the digits $6!$ and $9$ appears:
(a) $\dfrac{1}{2}\left( 6! \right)$
(b) ${{5}^{6}}$
(c) $\dfrac{5}{2}\left( 6! \right)$
(d) $6!$

Answer 1

Hint: In this question, we will first the number of ways of selecting repeated digits using combination and then apply permutation to find the answer.

Complete step-by-step answer:p answer:
In a given question, we are to form a 6-digit number, in which all and only $6!$ and $9$ appears.
We are to form 6-digit numbers with 5-digits, so in all cases, one of the digits will repeat twice and all other digits will appear once.
So, the number of ways in which we can choose 1 digit out of 5-digits to be repeated is given by $^{5}{{c}_{1}}$.
Now formula of combination
$^{n}{{c}_{r}}=\dfrac{n!}{\left( n-n \right)!n!}$
Therefore,
 $\begin{align}
  & ^{5}{{c}_{1}}=\dfrac{5!}{1!\left( 5-1 \right)!} \\
 & =\dfrac{5\times 4!}{4!}=5 \\
\end{align}$
So, we can choose the digits to be repeated in 5 ways.
Now, since, this number is repeated twice in the required number. So, it will be divided by 2 to find the number of such numbers. Also, for 6-digit numbers, with all the digits already selected, the number of ways in which we can form a number is $6!$.
Therefore, the number of 6-digits numbers with 5-digits $6!$ and $9$ is
 $6!\times \dfrac{^{5}{{c}_{1}}}{2}=\dfrac{5}{2}\times 6!$
Hence the correct answer is option (c).

Note: In this type of question, when digits are to be repeated on digits to be used are unsure, we first apply combinations to select digits and then apply permutation.