How many numbers lying between 10 and 1000 can be formed with the digits 2 3 4 0 and 7 when no digit is repeated?

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Related The number of numbers lying between 10 and 1000 can be formed with the digits 2,3,4,0,8,9 isa)124b)120c)125d)none of theseCorrect answer is option 'C'. Can you explain this answer?

Case 1:  two digit numbers:

We can choose the tens digit any of 5 ways (0 cannot be chosen)

We can then choose the units digit any of 5 ways, because 0 can be chosen.

That's 5 x 5 or 25 two digit numbers

Case 2:  three digit numbers:

We can choose the hundreds digit any of 5 ways (0 cannot be chosen)

We can then choose the tens digit any of 5 ways, because 0 can be chosen.

We can then choose the units digit any of the 4 remaining ways.

That's 5 x 5 x 4 = 100 three-digit numbers

So, total: 25+100 = 125

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Related The number of numbers lying between 10 and 1000 can be formed with the digits 2,3,4,0,8,9 isa)124b)120c)125d)none of theseCorrect answer is option 'C'. Can you explain this answer?

Case 1 :5*5=25(excluding zero for fist place)case 2:5*5*4=100(same excluding zero st first)total numbers=25+100=125

option (C) 125

1st digit could be chosen in 5 ways (excluding 0)

2nd digit could be chosen in 5 ways again (though the digit chosen earlier can’t be repeated, we have zero (0) in lieu of the utilized digit)

Thus we have 5*5 = 25 (Two digit Numbers meeting the criteria)

For 1st digit, we have 5 choices

For 2nd digit, we have 5 choices

For 3rd digit, we have 4 choices

Thus we have 5*5*4 = 100 (Three digit Numbers meeting the criteria)

So, we would have 25+100 = 125 Numbers meeting the required criteria.

125 is the required answer!

Related The number of numbers lying between 10 and 1000 can be formed with the digits 2,3,4,0,8,9 isa)124b)120c)125d)none of theseCorrect answer is option 'C'. Can you explain this answer?

Two digits number (0 can't come on unit place) 5 × 5 = 25Three digits number 5 × 5 × 4 = 100

Hence the total number of numbers between 10 and 1000 that can be formed using given digits is = 125

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Answer

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Hint: In these type of questions to find the number of numbers that lie between 10 and 1000 that can formed by using 0, 2, 3, 4, 5, 6 we have to first find the total two-digit numbers that lie between the numbers, and after that we have to find three digit numbers and add the number of two digit and number of three digit numbers to get the total number of numbers that lie between 10-1000.

Complete step-by-step solution:

Given range in the question is 10-1000 and we have to find number of numbers that lie between them and that can be formed by using the numbers 0, 2, 3, 4, 5, 6, and also given that the repetition of numbers is not allowed,Now first we have to find the number of two digit numbers, in first case 0 cannot be in ten’s place as it becomes single digit, Number of ways ten’s place in two digit numbers can be chosen using given numbers as 0 cannot chosen = 5,Number of ways unit’s place in two digit numbers can be chosen using given numbers as 0 can be chosen = 5,Number of two digit numbers that lie between 10-1000 =$5 \times 5$.$\therefore$ Number of two digit numbers that lie between 10-1000 = 25.Now we have to find the three digit numbers, In first case 0 cannot be in hundred’s place as it becomes double digit i.e., 056, 079, etc Number of ways hundred ’s place in three digit numbers can be chosen using given numbers as 0 cannot chosen = 5,Number of ways ten’s place in two digit numbers can be chosen using given numbers as 0 can be chosen = 5,Number of ways a unit's place in two digit numbers can be chosen using given numbers as 0 can be chosen = 4, as the numbers cannot be repeated. Number of two digit numbers that lie between 10-1000 =$5 \times 5 \times 4$.$\therefore$ Number of two digit numbers that lie between 10-1000 = 100.So, the total number of numbers that lie between 10-1000 that can be formed using numbers 0, 2, 3, 4, 5, 6 = number of two-digit numbers + number of three digit numbers.The total numbers = 100 +25.$\therefore$The total number of numbers = 125.

Option B is the correct answer.

Note: In these type of permutations questions , we use the method of counting, if we get $x$ ways to do then first thing, and if we get $y$ ways do the second thing then the total number of ways will become $x \times y$ ways, and this counting continuous till we get the required result and that depends on the given question.