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11 Feb 2021, 03:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
60% (02:58) correct
40%
(02:59)
wrong
based on 100
sessions
History
Date
Time
Result
Not Attempted Yet
Sameer has to make a telephone call to his friend Harish, 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. The maximum number of trials that Sameer has to make to be successful is
(A) 10,000
(B) 5000
(C) 3402
(D) 3200
(E) 3000
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 10,000
(B) 5000
(C) 3402
(D) 3200
(E) 3000
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
12 Feb 2021, 01:23
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Case 1:
When the last digit is 9.
First 3 digits = \(\binom{635}{674}\) or, No. of ways = \(2\) ways
Last digit = \(9\), or, No of ways = \(1\) way
No. of ways to fill the remaining 3 digits each =\(9 (1, 2,\cdots 9)\) ways
Thus, the total no. of ways when the last digit is \(9 = 2 {\tiny\text{(First 3 digits)}} \times 9 {\tiny\text{(4th position)}} \times 9 {\tiny\text{(5th position)}} \times 9 {\tiny\text{(6th position)}} \times 1{\tiny\text{(Last digit is 9)}} = 729 \times 2\)
Case 2:
When 9 is at any other position except for the last position, then it can be placed in the next 3 positions (i.e. \(4^{th}, 5^{th}, 6^{th}\) place), and the last position be \(\{1,3,5,7\}\)
Therefore, no. of ways \(= 2 {\tiny\text{(First 3 digits)}} \times 9 \times 9 \times 3 {\tiny\text{(when position of 9 changes)}} \times 4 {\tiny\text{(Last digit)}} = 972 \times 2\)
Hence, the total no of ways \(= 1458 + 1944 =3402\)
Ans C
When the last digit is 9.
First 3 digits = \(\binom{635}{674}\) or, No. of ways = \(2\) ways
Last digit = \(9\), or, No of ways = \(1\) way
No. of ways to fill the remaining 3 digits each =\(9 (1, 2,\cdots 9)\) ways
Thus, the total no. of ways when the last digit is \(9 = 2 {\tiny\text{(First 3 digits)}} \times 9 {\tiny\text{(4th position)}} \times 9 {\tiny\text{(5th position)}} \times 9 {\tiny\text{(6th position)}} \times 1{\tiny\text{(Last digit is 9)}} = 729 \times 2\)
Case 2:
When 9 is at any other position except for the last position, then it can be placed in the next 3 positions (i.e. \(4^{th}, 5^{th}, 6^{th}\) place), and the last position be \(\{1,3,5,7\}\)
Therefore, no. of ways \(= 2 {\tiny\text{(First 3 digits)}} \times 9 \times 9 \times 3 {\tiny\text{(when position of 9 changes)}} \times 4 {\tiny\text{(Last digit)}} = 972 \times 2\)
Hence, the total no of ways \(= 1458 + 1944 =3402\)
Ans C
