If C(16,r) = C(16,r+2), then find r. Explain how you know.
up vote
1
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Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?
combinations
asked 52 mins ago
Gold Pony Boy
102
add a comment |
up vote
1
down vote
favorite
Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?
combinations
asked 52 mins ago
Gold Pony Boy
102
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?
combinations
asked 52 mins ago
Gold Pony Boy
102
Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?
combinations
combinations
asked 52 mins ago
Gold Pony Boy
102
asked 52 mins ago
Gold Pony Boy
102
asked 52 mins ago
Gold Pony Boy
102
asked 52 mins ago
Gold Pony Boy
102
asked 52 mins ago
Gold Pony Boy
102
102
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3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Note that $binom{n}{k} = binom{n}{n-k}.$
What equation does this give for $r$ in your scenario? What is the solution to that equation?
Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can only taken on at most twice is taken as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.
answered 47 mins ago
Display name
779313
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
23 mins ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
16 mins ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
14 mins ago
add a comment |
up vote
2
down vote
Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?
answered 46 mins ago
Carl Schildkraut
10.6k11438
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
30 mins ago
add a comment |
up vote
0
down vote
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
This gives yoy that $r=7$
answered 34 mins ago
Fareed AF
36711
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
9 mins ago
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
8 mins ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
4 mins ago
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Note that $binom{n}{k} = binom{n}{n-k}.$
What equation does this give for $r$ in your scenario? What is the solution to that equation?
Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can only taken on at most twice is taken as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.
answered 47 mins ago
Display name
779313
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
23 mins ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
16 mins ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
14 mins ago
add a comment |
up vote
3
down vote
accepted
Note that $binom{n}{k} = binom{n}{n-k}.$
What equation does this give for $r$ in your scenario? What is the solution to that equation?
Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can only taken on at most twice is taken as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.
answered 47 mins ago
Display name
779313
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
23 mins ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
16 mins ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
14 mins ago
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Note that $binom{n}{k} = binom{n}{n-k}.$
What equation does this give for $r$ in your scenario? What is the solution to that equation?
Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can only taken on at most twice is taken as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.
answered 47 mins ago
Display name
779313
Note that $binom{n}{k} = binom{n}{n-k}.$
What equation does this give for $r$ in your scenario? What is the solution to that equation?
Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can only taken on at most twice is taken as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.
answered 47 mins ago
Display name
779313
edited 13 mins ago
answered 47 mins ago
Display name
779313
answered 47 mins ago
Display name
779313
answered 47 mins ago
Display name
779313
779313
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
23 mins ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
16 mins ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
14 mins ago
add a comment |
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
23 mins ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
16 mins ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
14 mins ago
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
23 mins ago
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
23 mins ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
16 mins ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
16 mins ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
14 mins ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
14 mins ago
add a comment |
up vote
2
down vote
Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?
answered 46 mins ago
Carl Schildkraut
10.6k11438
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
30 mins ago
add a comment |
up vote
2
down vote
Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?
answered 46 mins ago
Carl Schildkraut
10.6k11438
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
30 mins ago
add a comment |
up vote
2
down vote
up vote
2
down vote
Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?
answered 46 mins ago
Carl Schildkraut
10.6k11438
Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?
answered 46 mins ago
Carl Schildkraut
10.6k11438
answered 46 mins ago
Carl Schildkraut
10.6k11438
answered 46 mins ago
Carl Schildkraut
10.6k11438
answered 46 mins ago
Carl Schildkraut
10.6k11438
10.6k11438
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
30 mins ago
add a comment |
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
30 mins ago
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
30 mins ago
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
30 mins ago
add a comment |
up vote
0
down vote
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
This gives yoy that $r=7$
answered 34 mins ago
Fareed AF
36711
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
9 mins ago
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
8 mins ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
4 mins ago
add a comment |
up vote
0
down vote
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
This gives yoy that $r=7$
answered 34 mins ago
Fareed AF
36711
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
9 mins ago
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
8 mins ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
4 mins ago
add a comment |
up vote
0
down vote
up vote
0
down vote
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
This gives yoy that $r=7$
answered 34 mins ago
Fareed AF
36711
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
This gives yoy that $r=7$
answered 34 mins ago
Fareed AF
36711
answered 34 mins ago
Fareed AF
36711
answered 34 mins ago
Fareed AF
36711
answered 34 mins ago
Fareed AF
36711
36711
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
9 mins ago
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
8 mins ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
4 mins ago
add a comment |
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
9 mins ago
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
8 mins ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
4 mins ago
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
9 mins ago
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
9 mins ago
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
8 mins ago
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
8 mins ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
4 mins ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
4 mins ago
add a comment |
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