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The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,3,.9 such that each digit does not occur more than once in each number is (A) 150 (B) 152 (C) 154 (D) none Option (C) will be the answer. Please explain.?
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The total number of numbers less than 1000 and divisible by 5 formed w...
Solution:
Given that the numbers are divisible by 5 and are formed using digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 such that each digit occurs only once.
We need to find the number of such numbers less than 1000.
Let's consider the units place:
- The units place can be filled with either 0 or 5. So, there are 2 options for the units place.
Let's consider the tens place:
- If the units place is 0, then the tens place can be filled with any of the remaining 8 digits (1, 2, 3, 4, 6, 7, 8, 9). So, there are 8 options for the tens place.
- If the units place is 5, then the tens place can be filled with any of the remaining 7 digits (0, 1, 2, 3, 4, 6, 7, 8, 9). So, there are 7 options for the tens place.
Let's consider the hundreds place:
- If the tens place is filled with a digit other than 5, then the hundreds place can be filled with any of the remaining 6 digits (0, 1, 2, 3, 4, 6, 7, 8, 9). So, there are 6 options for the hundreds place.
- If the tens place is filled with 5, then the hundreds place can be filled with any of the remaining 5 digits (0, 1, 2, 3, 4, 6, 7, 8, 9). So, there are 5 options for the hundreds place.
Therefore, the total number of such numbers less than 1000 is:
2 × (8 + 7) × (6 + 5) = 154
Hence, the answer is (C) 154.
Given that the numbers are divisible by 5 and are formed using digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 such that each digit occurs only once.
We need to find the number of such numbers less than 1000.
Let's consider the units place:
- The units place can be filled with either 0 or 5. So, there are 2 options for the units place.
Let's consider the tens place:
- If the units place is 0, then the tens place can be filled with any of the remaining 8 digits (1, 2, 3, 4, 6, 7, 8, 9). So, there are 8 options for the tens place.
- If the units place is 5, then the tens place can be filled with any of the remaining 7 digits (0, 1, 2, 3, 4, 6, 7, 8, 9). So, there are 7 options for the tens place.
Let's consider the hundreds place:
- If the tens place is filled with a digit other than 5, then the hundreds place can be filled with any of the remaining 6 digits (0, 1, 2, 3, 4, 6, 7, 8, 9). So, there are 6 options for the hundreds place.
- If the tens place is filled with 5, then the hundreds place can be filled with any of the remaining 5 digits (0, 1, 2, 3, 4, 6, 7, 8, 9). So, there are 5 options for the hundreds place.
Therefore, the total number of such numbers less than 1000 is:
2 × (8 + 7) × (6 + 5) = 154
Hence, the answer is (C) 154.
Community Answer
The total number of numbers less than 1000 and divisible by 5 formed w...
Less than 1000 , so I consists of 1, 2 and 3 digit number
divisible by 5 , digits involved 0,1,2,3,....9
one digit number 5- one possibility
divisible by five so units place must be 0 or 5
two digit number _ _ ,
while calculating possibility in left corner zero should not be considered as if two digit number if we consider zero then it would make it as a one digit number
let's fix 0 in units place then 9*1 possibilities = 9
now let's fix 5 in units place so (10-1-1)*1
8*1= 8
total = 17
three digit number, _ _ _
hundred place 9 possibility as
fixing 0
9*8*1 = 72
fixing 5
8*8*1 = 64
total = 136
while calculating hundred place zero is excluded and in tens place zero is included
one digit = 1
two digit = 17
three digit = 136
total =154
so option C
divisible by 5 , digits involved 0,1,2,3,....9
one digit number 5- one possibility
divisible by five so units place must be 0 or 5
two digit number _ _ ,
while calculating possibility in left corner zero should not be considered as if two digit number if we consider zero then it would make it as a one digit number
let's fix 0 in units place then 9*1 possibilities = 9
now let's fix 5 in units place so (10-1-1)*1
8*1= 8
total = 17
three digit number, _ _ _
hundred place 9 possibility as
fixing 0
9*8*1 = 72
fixing 5
8*8*1 = 64
total = 136
while calculating hundred place zero is excluded and in tens place zero is included
one digit = 1
two digit = 17
three digit = 136
total =154
so option C
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The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,3,.9 such that each digit does not occur more than once in each number is (A) 150 (B) 152 (C) 154 (D) none Option (C) will be the answer. Please explain.?
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The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,3,.9 such that each digit does not occur more than once in each number is (A) 150 (B) 152 (C) 154 (D) none Option (C) will be the answer. Please explain.? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,3,.9 such that each digit does not occur more than once in each number is (A) 150 (B) 152 (C) 154 (D) none Option (C) will be the answer. Please explain.? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,3,.9 such that each digit does not occur more than once in each number is (A) 150 (B) 152 (C) 154 (D) none Option (C) will be the answer. Please explain.?.
The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,3,.9 such that each digit does not occur more than once in each number is (A) 150 (B) 152 (C) 154 (D) none Option (C) will be the answer. Please explain.? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,3,.9 such that each digit does not occur more than once in each number is (A) 150 (B) 152 (C) 154 (D) none Option (C) will be the answer. Please explain.? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,3,.9 such that each digit does not occur more than once in each number is (A) 150 (B) 152 (C) 154 (D) none Option (C) will be the answer. Please explain.?.
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The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,3,.9 such that each digit does not occur more than once in each number is (A) 150 (B) 152 (C) 154 (D) none Option (C) will be the answer. Please explain.?, a detailed solution for The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,3,.9 such that each digit does not occur more than once in each number is (A) 150 (B) 152 (C) 154 (D) none Option (C) will be the answer. Please explain.? has been provided alongside types of The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,3,.9 such that each digit does not occur more than once in each number is (A) 150 (B) 152 (C) 154 (D) none Option (C) will be the answer. Please explain.? theory, EduRev gives you an
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