Question

An old man while dialing a 7 digit telephone number remembers that the first four digit consists of one 1’s, one 2’s and two 3’s. He also remembers that the fifth digit is either a 4 or 5 while he has no memorization 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

Hint:In this question remember to use the 6 digit as x also use the concept of factorial to find the number of possible ways to arrange starting 4 digits and to find the 6th and 7th digit use the given information and find the possible combination of 6th and 7th digit this information will help you to find the total numbers of trails to dial a correct number.

Complete step-by-step answer:
According to the information given , the telephone number has 7 digits.
Now we know that the starting 4 digit dialed by the old man have one 1’s, one2’s and two 3’s
So the number of ways that we can arrange starting 4 digits number are = $ \dfrac{{4!}}{{2!}} $ = $ \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1}} $ = 12
So we can arrange the starting 4 digits number in 12 ways
Since we know that old man remember that 4 or 5 are the 5th digits
So by the above statement we can say that there are only two ways of dialing the 5th digit
For 6th digit since old man don’t remember the 6th digit but remember that the 7th digit is equals to 9 less than the 6th digit
So, let x be the 6th digit
Therefore 7th digit = 9 – x
Since we know that a telephone have numbers from 0 to 9 therefore the possible combination of 6th and 7th digits are:
(0, 9), (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0) so here are the 10 combinations that are possible for the (6th digit, 7th digit)
Hence the total number of trails to dial correct number dialed by old man are = $ 12 \times 2 \times 10 = 240 $
Therefore 240 are the maximum number of trails to dial the exact number
Hence, option B is the correct option.

Note: In the above question to find the number of ways to arrange the starting 4 digit we used the concept of factorial which can be explained as it is a function or an operation used in mathematical calculations in which all the whole number smaller than the given number is multiplied it is represented as n! Here “!” is the symbol of factorial and the solution of n! Is found as $ n! = n \times \left( {n - 1} \right) \times ...... \times 2 \times 1 $ .