Question

Sameer has to make a phone call to his friend Harish. Unfortunately, he does not remember the $ 7 $ digit phone number. But he remembers that the first three digits are either $ 635 $ or $ 674 $ .Also, the number is odd and has exactly one $ 9 $ in the number. The maximum number of trials that Sameer has to make to be successful is:
(A) $ 10,000 $
(B) $ 3,402 $
(C) $ 3,200 $
(D) $ 5,000 $

Answer 1

Hint: In the given question we are required to find out the number of maximum trials such that Sameer can figure out the number of his friend and can successfully talk to him given all the information in the problem itself. The phone number has seven digits and there are several conditions given in the question.

Complete step-by-step answer:
So, we are required to find the number of ways in which a seven digit phone number can be formed with the given conditions.
First, we know that the number is odd; there is exactly one $ 9 $ in the number.
Now, we have two possibilities. If the number ends with a $ 9 $ , then there is no other $ 9 $ in the number and if the number ends with any other odd digit except $ 9 $ , then there must be a $ 9 $ in the remaining places.
So, we would take both the cases separately.
Also, we are given that the first three digits of the phone number are either $ 635 $ or $ 674 $ . So, there are two options to fill up the first three digits of the phone number in both the cases.
Now, if the number ends with the digit $ 9 $ , then there is only way of filling up the last digit. But all the $ 3 $ remaining places of the number can be filled in $ 9 $ possible ways each.
Using the fundamental principle of counting, we get the total number of numbers obtained in this case as: $ 2 \times 9 \times 9 \times 9 \times 1 = 1458 $ .
Now, if the number does not end with the digit $ 9 $ , then the last digit can be filled with any odd number other than nine. So, there are $ 4 $ ways of filling up the last digit in this case. But one out of the $ 3 $ remaining places of the number has to be $ 9 $ . So, we have to first select which out of the three places would bear a nine.
We know that the number of ways of selecting a place out of three places is $ ^3{C_1} = 3 $ . Also, the remaining two positions can be filled with any of the digits except nine. So, we have $ 9 $ options each for filling up the rest of the two positions.
So, using the fundamental principle of counting, the total number of numbers obtained in this case are: $ 2 \times 3 \times 9 \times 9 \times 4 = 1944 $ .
Now, adding the number of ways in both the cases so as to get the maximum number of trials.
Hence, $ 1458 + 1944 = 3402 $ .
Therefore, the maximum number of trials that Sameer has to make to be successful is $ 3402 $ . Hence, option (B) is correct.
So, the correct answer is “Option B”.

Note: The question revolves around the concepts of Permutations and Combinations. One should know about the principle rule of counting or the multiplication rule. Care should be taken while handling the calculations. Calculations should be verified once so as to be sure of the answer.