$4$ digit number that are divided by $5$ given $1,2,3,4,5$
$begingroup$
Given $5$ distinct numbers $1,2,3,4,5$. without repetitions, find number of $4$ digit odd numbers.
For odd numbers, the last number must be odd $(1,3,5)$ and one number must be picked for the last number. Thus ${}_3C_1$.
The remaining $3$ front numbers, any other number can be picked except for one odd number. So with that, we have only $4$ options to choose from and we must pick $3$.
However, my answer booklet put it as ${}_4P_3$.
Therefore, ${}_4P_3 times {}_3C_1 = text{answer}$.
Why is it "$P$"? Permutations ($P$) means that the order matters. why does the order matters? I thought as long as the last digit is odd means the number is odd?
combinatorics
edited Nov 27 '18 at 11:22
Tianlalu
3,08621038
asked Nov 27 '18 at 10:09
ErikienErikien
494
$endgroup$
add a comment |
$begingroup$
Given $5$ distinct numbers $1,2,3,4,5$. without repetitions, find number of $4$ digit odd numbers.
For odd numbers, the last number must be odd $(1,3,5)$ and one number must be picked for the last number. Thus ${}_3C_1$.
The remaining $3$ front numbers, any other number can be picked except for one odd number. So with that, we have only $4$ options to choose from and we must pick $3$.
However, my answer booklet put it as ${}_4P_3$.
Therefore, ${}_4P_3 times {}_3C_1 = text{answer}$.
Why is it "$P$"? Permutations ($P$) means that the order matters. why does the order matters? I thought as long as the last digit is odd means the number is odd?
combinatorics
edited Nov 27 '18 at 11:22
Tianlalu
3,08621038
asked Nov 27 '18 at 10:09
ErikienErikien
494
$endgroup$
add a comment |
$begingroup$
Given $5$ distinct numbers $1,2,3,4,5$. without repetitions, find number of $4$ digit odd numbers.
For odd numbers, the last number must be odd $(1,3,5)$ and one number must be picked for the last number. Thus ${}_3C_1$.
The remaining $3$ front numbers, any other number can be picked except for one odd number. So with that, we have only $4$ options to choose from and we must pick $3$.
However, my answer booklet put it as ${}_4P_3$.
Therefore, ${}_4P_3 times {}_3C_1 = text{answer}$.
Why is it "$P$"? Permutations ($P$) means that the order matters. why does the order matters? I thought as long as the last digit is odd means the number is odd?
combinatorics
edited Nov 27 '18 at 11:22
Tianlalu
3,08621038
asked Nov 27 '18 at 10:09
ErikienErikien
494
$endgroup$
Given $5$ distinct numbers $1,2,3,4,5$. without repetitions, find number of $4$ digit odd numbers.
For odd numbers, the last number must be odd $(1,3,5)$ and one number must be picked for the last number. Thus ${}_3C_1$.
The remaining $3$ front numbers, any other number can be picked except for one odd number. So with that, we have only $4$ options to choose from and we must pick $3$.
However, my answer booklet put it as ${}_4P_3$.
Therefore, ${}_4P_3 times {}_3C_1 = text{answer}$.
Why is it "$P$"? Permutations ($P$) means that the order matters. why does the order matters? I thought as long as the last digit is odd means the number is odd?
combinatorics
combinatorics
edited Nov 27 '18 at 11:22
Tianlalu
3,08621038
asked Nov 27 '18 at 10:09
ErikienErikien
494
edited Nov 27 '18 at 11:22
Tianlalu
3,08621038
asked Nov 27 '18 at 10:09
ErikienErikien
494
edited Nov 27 '18 at 11:22
Tianlalu
3,08621038
edited Nov 27 '18 at 11:22
Tianlalu
3,08621038
edited Nov 27 '18 at 11:22
Tianlalu
3,08621038
3,08621038
asked Nov 27 '18 at 10:09
ErikienErikien
494
asked Nov 27 '18 at 10:09
ErikienErikien
494
asked Nov 27 '18 at 10:09
ErikienErikien
494
494
add a comment |
add a comment |
1 Answer
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$begingroup$
The order matters because the numbers $2345$ and $4235$ are different numbers.
answered Nov 27 '18 at 10:18
5xum5xum
90.2k394161
$endgroup$
add a comment |
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$begingroup$
The order matters because the numbers $2345$ and $4235$ are different numbers.
answered Nov 27 '18 at 10:18
5xum5xum
90.2k394161
$endgroup$
add a comment |
$begingroup$
The order matters because the numbers $2345$ and $4235$ are different numbers.
answered Nov 27 '18 at 10:18
5xum5xum
90.2k394161
$endgroup$
add a comment |
$begingroup$
The order matters because the numbers $2345$ and $4235$ are different numbers.
answered Nov 27 '18 at 10:18
5xum5xum
90.2k394161
$endgroup$
The order matters because the numbers $2345$ and $4235$ are different numbers.
answered Nov 27 '18 at 10:18
5xum5xum
90.2k394161
answered Nov 27 '18 at 10:18
5xum5xum
90.2k394161
answered Nov 27 '18 at 10:18
5xum5xum
90.2k394161
answered Nov 27 '18 at 10:18
5xum5xum
90.2k394161
90.2k394161
add a comment |
add a comment |
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