Review How many odd numbers less than 1000 can be formed using the digits 0 3 5 7 when repetition of digits is not allowed?
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Nội dung chính- How many odd numbers less than 1000 can be formed using the digits 0 2 5 7 repetition of digits are allowed?How many odd numbers less than 1000 can be formed by using the digits 0 3 6 and 9?How many odd numbers less than 10000 can be formed using the digits 0 2 3 5 allowing repetition of digits?How many even numbers less than 1000 can be formed by using the digits 2 4 3 and 5 if repetition of the digits is allowed?
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Q.. How many odd numbers less than $1000$ can be formed by using the digits $0,3,5,7$. Repetition not allowed.
A. $21$
Answer is correct (please provide a thorough explanation).
Unit digit nos. : $3$
Dual digit nos. : $2×3$
Three digit nos. : $2×2×3$ (as per the answer)
Three digit nos. : $1×3×3$ (as per my viewpoint)
Please help.
Text Solution
Solution : Since
each number is less than 1000, required numbers are the 1-digit, 2-digit and 3-digit numbers.
One-digit numbers: Clearly, there are two one -digit odd numbers, namely 5 and 7, formed of the given digits.
Two-digit numbers: Since we are to form 2- digit odd numbers, we may put 5 or 7 the unit's place. So, there are 2 ways of filling the unit's place.
Now, we cannot use 0 the ten's place and the repetition of digits is allowed. So, we may fill up the ten's
place by any of the digits 2, 5, 7. Thus, there are 3 ways of filling the ten's place.
Hence, the required type of 2-digit numbers `=(2xx3)=6.`
Three-digit numbers: To have an odd 3-digit number, we may put 5 or 7 the unit's place. So, there are 2 ways of filling the unit's place.
We may fill up the ten's place by any of the digits 0, 2, 5, 7. So, there are 4 ways of filling the ten's place.
We cannot put 0 the hundred's place. So, the hundred's
place can be filled by any of the digits 2, 5, 7 and so it can be done in 3 ways.
`therefore " the required number of 3-digit numbers"= (2xx4xx3) =24.`
Hence, the total number of required type of numbers `=(2+6+24)= 32.`
Since the number is less than 1000, it could be a three-digit, two-digit or single-digit number.
Case I: Three-digit number:
Now, the hundred's place cannot be zero. Thus, it can be filled with three digits, i.e. 3, 5
and 7.Also, the unit's place cannot be zero. This is because it is an odd number and one digit has already been used to fill the hundred's place.
Thus, the unit's place can be filled by only 2 digits.
Number of ways of filling the ten's digit = 2 (as repetition is not allowed)
Total three-digit numbers that can be formed = `3xx2xx2=12`
Case II: Two-digit number:
Now, the ten's place cannot be zero. Thus, it can be
filled with three digits, i.e. 3, 5 and 7.Also, the unit's place cannot be zero. This is because it is an odd number and one digit has already been used to fill the ten's place,
Thus, the unit's place can be filled by only 2 digits.
Total two-digit numbers that can be formed = `3xx2=6`
Case III: Single-digit number: It could be 3, 5 and 7.Total single-digit numbers that can be formed = 3
Hence, required number = 12 + 6 + 3 = 21
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Q.. How many odd numbers less than $1000$ can be formed by using the digits $0,3,5,7$. Repetition not allowed.
A. $21$
Answer is correct (please provide a thorough explanation).
Unit digit nos. : $3$
Dual digit nos. : $2×3$
Three digit nos. : $2×2×3$ (as per the answer)
Three digit nos. : $1×3×3$ (as per my viewpoint)
Please help.
asked Jun 15, 2015 10:19
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For the three-digit number case, there are three ways of selecting the units digit since we cannot use $0$. Once the units digit has been chosen, there are two choices for the hundreds digit since we cannot use either zero or the units digit. That leaves us with two choices for the tens digit since we cannot use the hundreds digit or the units digit. Hence, the number of three-digit odd numbers that can be formed using only the digits $0, 3, 5, 7$ is $2 cdot 2 cdot 3$.
answered Jun 15, 2015 10:38
N. F. TaussigN. F. Taussig
68k13 gold badges52 silver badges70 bronze badges
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As I understand it, you think like the following: number xyz, 3 choices for x (3,5 or 7), then there are 3 choises for y (the two remaining odd numbers and 0) and then for choosing z there is only one choise, since it needs to be odd. Thus we have $3cdot 3cdot 1$ different numbers.
This is not correct since if $y=0$ then we have two different choises for the number $z$ not 1. Thus we may do callculations like this:
If $y=0$ then we have $3cdot 1 cdot 2$ choises.
If $yneq 0$ then we have $3cdot 2cdot 1$ choises.
Thus in total we have $3cdot 2cdot 1 + 3cdot 1cdot 2 = 3 cdot 2 cdot 2$ choises.
answered Jun 15, 2015 10:51
Ove AhlmanOve Ahlman
4,2611 gold badge14 silver badges27 bronze badges
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How many odd numbers less than 1000 can be formed using the digits 0 2 5 7 repetition of digits are allowed?
Hence, the odd numbers less than 1000 formed by using the digits 0, 2, 5 and 7 when the repetition of digits is allowed is given by 32 ways.How many odd numbers less than 1000 can be formed by using the digits 0 3 6 and 9?
Hence, total number of odd numbers less than 1000=24+6+2=32.How many odd numbers less than 10000 can be formed using the digits 0 2 3 5 allowing repetition of digits?
Answer: 128 odd numbers can be formed.How many even numbers less than 1000 can be formed by using the digits 2 4 3 and 5 if repetition of the digits is allowed?
Hence 42 even numbers less than 1000 can be formed by using the digits 2,4,3 and 5 if repetition to the digits is allowed. Tải thêm tài liệu liên quan đến nội dung bài viết How many odd numbers less than 1000 can be formed using the digits 0 3 5 7 when repetition of digits is not allowed?
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