Question

Sameer has to make a telephone call to his friend Nisheeth, Unfortunately he does not remember the 7 digit phone number. But he remembers that the first three digits are 635 or 674, the number is odd and there is exactly one 9 in the number. Find the maximum number of trials that Sameer has to make to be successful. \[\]
A.10000\[\]
B.3402\[\]
C.3200\[\]
D. 5000\[\]

Answer 1

Hint: We can either fill the first three places by 635 or 674. When we fill with 635 we have two cases either 9 is fixed in units or ${{7}^{\text{th}}}$ place or it is not. We use the formula that we can fill $n$distinct objects in $r$ places with repetition in ${{n}^{r}}$ways, the rule of sum and product to find the number of trials with first three digits 635 which will be exactly same for first three digits as 674. We add a number of trials for 635 and 674 to get the result.

Complete step-by-step answer:
We are given the question that Sameer has to make a telephone call to his friend Nisheeth, but he does not remember the 7 digit phone number. So we have 7 places to fill with digits from 0 to 9. \[\]
We are further given that he remembers that the first three digits are 635 or 674, the number is odd and there is exactly one 9 in the number. As the number is odd, there can be odd digits 1, 3, 5, 7, 9 and in the unit place of the phone number. \[\]
Let us fill the first three places by 635. We have to fill 4 vacant places. We have two cases here,
\[\begin{matrix}
   6 & 3 & 5 & \_ & \_ & \_ & \_ \\
   {} & {} & {} & {{100}^{\operatorname{th}}} & {{100}^{\text{th}}} & {{10}^{\text{th}}} & \text{unit} \\
\end{matrix}\]
 Case-1: We fix the one 9 in the unit place or ${{7}^{\text{th}}}$ place from left in 1 way. We can fill the rest three vacant places with 9 digits from 0 to 8 (since there can be only one 9) using the rule of product in $9\times 9\times 9\times 1=729$ ways. \[\]
Case-2: We do not fill the one 9 in the unit place. Since we can fill the unit place with only odd digits 1, 3, 5, 7 we have 4 ways. We can fix one 9 in vacant three places in ${}^{3}{{C}_{1}}=3$ ways. We can fill the rest two vacant places with 9 digits from 0 to 8 in $9\times 9$ ways. So by rule of product number ways for case-2 is $4\times 3\times 9\times 9=972$.\[\]
So by rule of sum total number of phone numbers with first three digits 635 is $729+972=1701$.Similarly we can fill the first three digits 674 and find the number of numbers 1701. We use rule of sum to find maximum number of trials that Sameer has to make to be successful as $1701+1701=3402$. \[\]

Note: We note that if we can do one thing in $m$ ways and another thing in $n$ ways then by rule of sum we can do either of the things in $m+n$ ways and by rule of product both of the things simultaneously in $mn$ ways. We can arrange $n$n distinct objects in $r$ places without repetition in ${}^{n}{{P}_{r}}$ ways.