Solving equations using permutations and combinations [closed]
$begingroup$
I need to solve for $n$ if $$P^n_4=60times C^n_2$$I’ve gotten up to $$n(n-1)(n-2)(n-3)=30n(n-1)$$
combinatorics permutations
edited Jan 14 at 18:13
Shubham Johri
5,412818
asked Jan 14 at 17:53
Sarah Sarah
31
$endgroup$
closed as off-topic by darij grinberg, Namaste, Holo, user91500, José Carlos Santos Jan 15 at 9:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I need to solve for $n$ if $$P^n_4=60times C^n_2$$I’ve gotten up to $$n(n-1)(n-2)(n-3)=30n(n-1)$$
combinatorics permutations
edited Jan 14 at 18:13
Shubham Johri
5,412818
asked Jan 14 at 17:53
Sarah Sarah
31
$endgroup$
closed as off-topic by darij grinberg, Namaste, Holo, user91500, José Carlos Santos Jan 15 at 9:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
simplify n(n-1) on both sides, you get (n-2)(n-3)=30, which leads to solution n=8 (or n=-3, which i suppose makes no sense because you are working with n positive)
$endgroup$
– tommycautero
Jan 14 at 18:17
add a comment |
$begingroup$
I need to solve for $n$ if $$P^n_4=60times C^n_2$$I’ve gotten up to $$n(n-1)(n-2)(n-3)=30n(n-1)$$
combinatorics permutations
edited Jan 14 at 18:13
Shubham Johri
5,412818
asked Jan 14 at 17:53
Sarah Sarah
31
$endgroup$
I need to solve for $n$ if $$P^n_4=60times C^n_2$$I’ve gotten up to $$n(n-1)(n-2)(n-3)=30n(n-1)$$
combinatorics permutations
combinatorics permutations
edited Jan 14 at 18:13
Shubham Johri
5,412818
asked Jan 14 at 17:53
Sarah Sarah
31
edited Jan 14 at 18:13
Shubham Johri
5,412818
asked Jan 14 at 17:53
Sarah Sarah
31
edited Jan 14 at 18:13
Shubham Johri
5,412818
edited Jan 14 at 18:13
Shubham Johri
5,412818
edited Jan 14 at 18:13
Shubham Johri
5,412818
5,412818
asked Jan 14 at 17:53
Sarah Sarah
31
asked Jan 14 at 17:53
Sarah Sarah
31
asked Jan 14 at 17:53
Sarah Sarah
31
31
closed as off-topic by darij grinberg, Namaste, Holo, user91500, José Carlos Santos Jan 15 at 9:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by darij grinberg, Namaste, Holo, user91500, José Carlos Santos Jan 15 at 9:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
simplify n(n-1) on both sides, you get (n-2)(n-3)=30, which leads to solution n=8 (or n=-3, which i suppose makes no sense because you are working with n positive)
$endgroup$
– tommycautero
Jan 14 at 18:17
add a comment |
$begingroup$
simplify n(n-1) on both sides, you get (n-2)(n-3)=30, which leads to solution n=8 (or n=-3, which i suppose makes no sense because you are working with n positive)
$endgroup$
– tommycautero
Jan 14 at 18:17
$begingroup$
simplify n(n-1) on both sides, you get (n-2)(n-3)=30, which leads to solution n=8 (or n=-3, which i suppose makes no sense because you are working with n positive)
$endgroup$
– tommycautero
Jan 14 at 18:17
$begingroup$
simplify n(n-1) on both sides, you get (n-2)(n-3)=30, which leads to solution n=8 (or n=-3, which i suppose makes no sense because you are working with n positive)
$endgroup$
– tommycautero
Jan 14 at 18:17
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Now your equation can be simplified to
$n(n-1)[(n-2)(n-3)-30]=0$
which can be further simplified to
$n(n-1)[n^2-5n-24]=0$
From here we get following values
$n=-3,0,1,8$
Now since $ngeq4$
$n=8$ is your answer.
answered Jan 14 at 18:16
JimmyJimmy
30813
$endgroup$
add a comment |
$begingroup$
By eliminating common terms on both sides, you get $(n-2)(n-3)=30$, or $n^2-5n-24=0$. Positive solution is $n=8$.
answered Jan 14 at 18:18
herb steinbergherb steinberg
3,0982311
$endgroup$
add a comment |
$begingroup$
You are nearly there! You got$$n(n-1)(n-2)(n-3)=30n(n-1)$$Now shift all the terms to one side to get$$n(n-1)Big[(n-2)(n-3)-30Big]=0$$So $n$ could be $0,1$ or such that$$(n-2)(n-3)-30=0\implies n^2-5n-24=(n-8)(n+3)=0\therefore n=0,1,-3,8$$If you define $P^n_4=displaystylefrac{n!}{(n-4)!}$, you need $nge4$. This leaves you with the answer $n=8$.
answered Jan 14 at 18:19
Shubham JohriShubham Johri
5,412818
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Now your equation can be simplified to
$n(n-1)[(n-2)(n-3)-30]=0$
which can be further simplified to
$n(n-1)[n^2-5n-24]=0$
From here we get following values
$n=-3,0,1,8$
Now since $ngeq4$
$n=8$ is your answer.
answered Jan 14 at 18:16
JimmyJimmy
30813
$endgroup$
add a comment |
$begingroup$
Now your equation can be simplified to
$n(n-1)[(n-2)(n-3)-30]=0$
which can be further simplified to
$n(n-1)[n^2-5n-24]=0$
From here we get following values
$n=-3,0,1,8$
Now since $ngeq4$
$n=8$ is your answer.
answered Jan 14 at 18:16
JimmyJimmy
30813
$endgroup$
add a comment |
$begingroup$
Now your equation can be simplified to
$n(n-1)[(n-2)(n-3)-30]=0$
which can be further simplified to
$n(n-1)[n^2-5n-24]=0$
From here we get following values
$n=-3,0,1,8$
Now since $ngeq4$
$n=8$ is your answer.
answered Jan 14 at 18:16
JimmyJimmy
30813
$endgroup$
Now your equation can be simplified to
$n(n-1)[(n-2)(n-3)-30]=0$
which can be further simplified to
$n(n-1)[n^2-5n-24]=0$
From here we get following values
$n=-3,0,1,8$
Now since $ngeq4$
$n=8$ is your answer.
answered Jan 14 at 18:16
JimmyJimmy
30813
answered Jan 14 at 18:16
JimmyJimmy
30813
answered Jan 14 at 18:16
JimmyJimmy
30813
answered Jan 14 at 18:16
JimmyJimmy
30813
30813
add a comment |
add a comment |
$begingroup$
By eliminating common terms on both sides, you get $(n-2)(n-3)=30$, or $n^2-5n-24=0$. Positive solution is $n=8$.
answered Jan 14 at 18:18
herb steinbergherb steinberg
3,0982311
$endgroup$
add a comment |
$begingroup$
By eliminating common terms on both sides, you get $(n-2)(n-3)=30$, or $n^2-5n-24=0$. Positive solution is $n=8$.
answered Jan 14 at 18:18
herb steinbergherb steinberg
3,0982311
$endgroup$
add a comment |
$begingroup$
By eliminating common terms on both sides, you get $(n-2)(n-3)=30$, or $n^2-5n-24=0$. Positive solution is $n=8$.
answered Jan 14 at 18:18
herb steinbergherb steinberg
3,0982311
$endgroup$
By eliminating common terms on both sides, you get $(n-2)(n-3)=30$, or $n^2-5n-24=0$. Positive solution is $n=8$.
answered Jan 14 at 18:18
herb steinbergherb steinberg
3,0982311
answered Jan 14 at 18:18
herb steinbergherb steinberg
3,0982311
answered Jan 14 at 18:18
herb steinbergherb steinberg
3,0982311
answered Jan 14 at 18:18
herb steinbergherb steinberg
3,0982311
3,0982311
add a comment |
add a comment |
$begingroup$
You are nearly there! You got$$n(n-1)(n-2)(n-3)=30n(n-1)$$Now shift all the terms to one side to get$$n(n-1)Big[(n-2)(n-3)-30Big]=0$$So $n$ could be $0,1$ or such that$$(n-2)(n-3)-30=0\implies n^2-5n-24=(n-8)(n+3)=0\therefore n=0,1,-3,8$$If you define $P^n_4=displaystylefrac{n!}{(n-4)!}$, you need $nge4$. This leaves you with the answer $n=8$.
answered Jan 14 at 18:19
Shubham JohriShubham Johri
5,412818
$endgroup$
add a comment |
$begingroup$
You are nearly there! You got$$n(n-1)(n-2)(n-3)=30n(n-1)$$Now shift all the terms to one side to get$$n(n-1)Big[(n-2)(n-3)-30Big]=0$$So $n$ could be $0,1$ or such that$$(n-2)(n-3)-30=0\implies n^2-5n-24=(n-8)(n+3)=0\therefore n=0,1,-3,8$$If you define $P^n_4=displaystylefrac{n!}{(n-4)!}$, you need $nge4$. This leaves you with the answer $n=8$.
answered Jan 14 at 18:19
Shubham JohriShubham Johri
5,412818
$endgroup$
add a comment |
$begingroup$
You are nearly there! You got$$n(n-1)(n-2)(n-3)=30n(n-1)$$Now shift all the terms to one side to get$$n(n-1)Big[(n-2)(n-3)-30Big]=0$$So $n$ could be $0,1$ or such that$$(n-2)(n-3)-30=0\implies n^2-5n-24=(n-8)(n+3)=0\therefore n=0,1,-3,8$$If you define $P^n_4=displaystylefrac{n!}{(n-4)!}$, you need $nge4$. This leaves you with the answer $n=8$.
answered Jan 14 at 18:19
Shubham JohriShubham Johri
5,412818
$endgroup$
You are nearly there! You got$$n(n-1)(n-2)(n-3)=30n(n-1)$$Now shift all the terms to one side to get$$n(n-1)Big[(n-2)(n-3)-30Big]=0$$So $n$ could be $0,1$ or such that$$(n-2)(n-3)-30=0\implies n^2-5n-24=(n-8)(n+3)=0\therefore n=0,1,-3,8$$If you define $P^n_4=displaystylefrac{n!}{(n-4)!}$, you need $nge4$. This leaves you with the answer $n=8$.
answered Jan 14 at 18:19
Shubham JohriShubham Johri
5,412818
answered Jan 14 at 18:19
Shubham JohriShubham Johri
5,412818
answered Jan 14 at 18:19
Shubham JohriShubham Johri
5,412818
answered Jan 14 at 18:19
Shubham JohriShubham Johri
5,412818
5,412818
add a comment |
add a comment |
$begingroup$
simplify n(n-1) on both sides, you get (n-2)(n-3)=30, which leads to solution n=8 (or n=-3, which i suppose makes no sense because you are working with n positive)
$endgroup$
– tommycautero
Jan 14 at 18:17