Question

An old man while dialing a $7$ digit telephone number remembers that the first four digits consists of one $1^{\prime }s$, one $2^{\prime }s$ and two $3^{\prime }s$. He also remembers that the fifth digits is either a $4$ or $5$ while has no memorising of the sixth digit, he remembers that the seventh digit is $9$ minus the sixth digit. Maximum number of distinct trials he has to try to make sure that he dials the correct telephone number, is

• A. $360$
• B. $240$
• C. $216$
• D. None of these

Answer 1

The correct option is B $240$

The telephone number has $7$ digits.

The first $4$ digits have one $1$, one $2$ and $2$ $3$'s.

$\therefore$ No. of ways of arranging the first $4$ digits$=\frac{4!}{2!}=12$ ways

The $5^{th}$ digit is a $4$ or a $5$.

$\therefore$ There are $2$ options for the $5^{th}$ digit.

Let the $6^{th}$ digit be x. Then, the $7^{th}$ digit is $9-x$.

$\therefore$ Possible combinations for $6^{th}$ and $7^{th}$ digits are:

($6^{th}$ digit, $7^{th}$ digit): $(0,9),(1,8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1),(9,0)$.

$\therefore$ There are $10$ possible combinations for the $6^{th}$ and $7^{th}$ digits.

$\therefore$ No. of possible telephone numbers$=12\times 2\times 10=240$, to dial correct number.

$\therefore$ The correct answer is option (B).