Ex 6 1 1 How Many 3 Digit Numbers From 1 2о Ex 6.1, 1how many 3 digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that(ii) repetition of the digits is not allowed?3 digit number : number of. First, assume that the ones place has a 0. then, the hundreds place can have any of {1,2,3,4,5,6}, so there are six choices. following this, the tens place can have any of the remaining five. next, assume the ones place has a 2, 4, or 6. now, the hundreds place has five choices, since both 0 and one more number are removed.
Solved How Many 3 Digit Numbers Can Be Formed Using The Digits 1 2о How many $3$ digit numbers can be formed by $1,2,3,4$, when the repetition of digits is allowed? so basically, i attempted this question as there are 4 numbers and 3 places to put in the numbers: in the ones place, any 4 numbers can be put, so there are 4 choices in the ones place. similarly for the tens and the hundreds place. Using basics it would be $ 5 \times 5 \times4 = 100$ 2) how many 3 digit numbers can be formed by $8,1,2,3,4,5$ which are even? again using basics we get $ 4 \times 5 \times 3 =60$ 3) now i want to ask how many 3 digit numbers can be formed which are even using $0,1,2,3,4,5$? no repetition is allowed in all above cases. Since, repetition is allowed , so tens place can also be filled by 6 ways. similarly,hundreds place can also be filled by 6 ways. so, number of ways in which three digit even numbers can be formed from the given digits is 6 × 6 × 3 = 108. The given digits are 1, 2, 3, 4, 5 and 6. there are many ways to form 3 digit even numbers from the digits 1, 2, 3, 4, 5 and 6. since even numbers contain even digits.
How Many 3 Digit Even Numbers Can Be Formed From The Digits 1 ођ Since, repetition is allowed , so tens place can also be filled by 6 ways. similarly,hundreds place can also be filled by 6 ways. so, number of ways in which three digit even numbers can be formed from the given digits is 6 × 6 × 3 = 108. The given digits are 1, 2, 3, 4, 5 and 6. there are many ways to form 3 digit even numbers from the digits 1, 2, 3, 4, 5 and 6. since even numbers contain even digits. That leaves 5 unchosen digits including 0. choose the middle digit any of these 5 unchosen ways. answer: (3)(5)(5) = 75 ways (iii) how many are greater than 330 ? case 1: all those between 330 and 400 choose the first digit 1 way (as 3) that leaves 6 unchosen digits. choose the second digit any of 3 ways, 4,5, or 6 that leaves 5 unchosen digits. So,required number of ways in which three digit even numbers can be formed from the given digits is 4×5×3 = 60. alternative method: 3 digit even numbers are to be formed using the given six digits, ,2,3,4,6 and 7, without repeating the digits. then, units digits can be filled in 3 ways by any of the digits, 2,4 or 6.
Ex 6 3 1 How Many 3 Digit Numbers Can Be Formed By U That leaves 5 unchosen digits including 0. choose the middle digit any of these 5 unchosen ways. answer: (3)(5)(5) = 75 ways (iii) how many are greater than 330 ? case 1: all those between 330 and 400 choose the first digit 1 way (as 3) that leaves 6 unchosen digits. choose the second digit any of 3 ways, 4,5, or 6 that leaves 5 unchosen digits. So,required number of ways in which three digit even numbers can be formed from the given digits is 4×5×3 = 60. alternative method: 3 digit even numbers are to be formed using the given six digits, ,2,3,4,6 and 7, without repeating the digits. then, units digits can be filled in 3 ways by any of the digits, 2,4 or 6.
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