Counting $k$-tuples from non-distinct collection of elements.
0
Assume that we have a collection of objects ${n_1,n_1,n_2,n_2,...,n_k,n_k}$. In how many ways can we construct a set ${n_1,...,n_k}$ consisting of distinct $k$-tuples?
combinatorics
asked Dec 7 at 8:01
user430191
1799
add a comment |
0
Assume that we have a collection of objects ${n_1,n_1,n_2,n_2,...,n_k,n_k}$. In how many ways can we construct a set ${n_1,...,n_k}$ consisting of distinct $k$-tuples?
combinatorics
asked Dec 7 at 8:01
user430191
1799
add a comment |
0
0
0
Assume that we have a collection of objects ${n_1,n_1,n_2,n_2,...,n_k,n_k}$. In how many ways can we construct a set ${n_1,...,n_k}$ consisting of distinct $k$-tuples?
combinatorics
asked Dec 7 at 8:01
user430191
1799
Assume that we have a collection of objects ${n_1,n_1,n_2,n_2,...,n_k,n_k}$. In how many ways can we construct a set ${n_1,...,n_k}$ consisting of distinct $k$-tuples?
combinatorics
combinatorics
asked Dec 7 at 8:01
user430191
1799
asked Dec 7 at 8:01
user430191
1799
asked Dec 7 at 8:01
user430191
1799
asked Dec 7 at 8:01
user430191
1799
asked Dec 7 at 8:01
user430191
1799
1799
add a comment |
add a comment |
1 Answer
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You must choose one from each two-element multiset ${n_i, n_i}$, and you must choose $k$ times. Thus there are $2^k$ ways in total.
answered Dec 7 at 8:06
William Sun
440111
Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
– user430191
Dec 7 at 8:14
I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
– William Sun
Dec 7 at 8:19
I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
– user430191
Dec 7 at 8:21
@user430191 Then clearly there are only two ways, since only $n_j$ can alter.
– William Sun
Dec 7 at 8:23
Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
– user430191
Dec 7 at 8:28
|
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1 Answer
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1 Answer
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You must choose one from each two-element multiset ${n_i, n_i}$, and you must choose $k$ times. Thus there are $2^k$ ways in total.
answered Dec 7 at 8:06
William Sun
440111
Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
– user430191
Dec 7 at 8:14
I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
– William Sun
Dec 7 at 8:19
I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
– user430191
Dec 7 at 8:21
@user430191 Then clearly there are only two ways, since only $n_j$ can alter.
– William Sun
Dec 7 at 8:23
Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
– user430191
Dec 7 at 8:28
|
show 1 more comment
1
You must choose one from each two-element multiset ${n_i, n_i}$, and you must choose $k$ times. Thus there are $2^k$ ways in total.
answered Dec 7 at 8:06
William Sun
440111
Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
– user430191
Dec 7 at 8:14
I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
– William Sun
Dec 7 at 8:19
I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
– user430191
Dec 7 at 8:21
@user430191 Then clearly there are only two ways, since only $n_j$ can alter.
– William Sun
Dec 7 at 8:23
Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
– user430191
Dec 7 at 8:28
|
show 1 more comment
1
1
1
You must choose one from each two-element multiset ${n_i, n_i}$, and you must choose $k$ times. Thus there are $2^k$ ways in total.
answered Dec 7 at 8:06
William Sun
440111
You must choose one from each two-element multiset ${n_i, n_i}$, and you must choose $k$ times. Thus there are $2^k$ ways in total.
answered Dec 7 at 8:06
William Sun
440111
answered Dec 7 at 8:06
William Sun
440111
answered Dec 7 at 8:06
William Sun
440111
answered Dec 7 at 8:06
William Sun
440111
440111
Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
– user430191
Dec 7 at 8:14
I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
– William Sun
Dec 7 at 8:19
I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
– user430191
Dec 7 at 8:21
@user430191 Then clearly there are only two ways, since only $n_j$ can alter.
– William Sun
Dec 7 at 8:23
Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
– user430191
Dec 7 at 8:28
|
show 1 more comment
Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
– user430191
Dec 7 at 8:14
I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
– William Sun
Dec 7 at 8:19
I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
– user430191
Dec 7 at 8:21
@user430191 Then clearly there are only two ways, since only $n_j$ can alter.
– William Sun
Dec 7 at 8:23
Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
– user430191
Dec 7 at 8:28
Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
– user430191
Dec 7 at 8:14
Thank you very much for the answer. Thus similarly, if we want for instance a collection where two of the $n_i$'s are equal only, then we have $2^{k-1}$ ways right?
– user430191
Dec 7 at 8:14
I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
– William Sun
Dec 7 at 8:19
I don’t quite understand what you mean. If only two $n_i$’s are equal there are only two ways to select. If only two are non-equal then we would have $2^{k-1}$ ways.
– William Sun
Dec 7 at 8:19
I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
– user430191
Dec 7 at 8:21
I mean the number of ways, a collection like ${n_1,...,n_j,n_j,....,n_k}$ where all but the $n_j$'s are distinct.
– user430191
Dec 7 at 8:21
@user430191 Then clearly there are only two ways, since only $n_j$ can alter.
– William Sun
Dec 7 at 8:23
@user430191 Then clearly there are only two ways, since only $n_j$ can alter.
– William Sun
Dec 7 at 8:23
Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
– user430191
Dec 7 at 8:28
Perhaps I don't phrase that correctly; say that we choose from the beginning that both $n_j$'s are in the collection and we want all the other $k-2$ elements being distinct. In how many ways can we construct such a set this time?
– user430191
Dec 7 at 8:28
|
show 1 more comment
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