If $C(16,r) = C(16,r+2)$, then find $r$. Explain how you know.
up vote
2
down vote
favorite
If $C(16,r) = C(16,r+2)$, then find $r$. Explain how you know.
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
edited Dec 3 at 9:23
Tianlalu
2,9801936
asked Dec 3 at 4:23
Gold Pony Boy
152
add a comment |
up vote
2
down vote
favorite
If $C(16,r) = C(16,r+2)$, then find $r$. Explain how you know.
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
edited Dec 3 at 9:23
Tianlalu
2,9801936
asked Dec 3 at 4:23
Gold Pony Boy
152
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If $C(16,r) = C(16,r+2)$, then find $r$. Explain how you know.
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
edited Dec 3 at 9:23
Tianlalu
2,9801936
asked Dec 3 at 4:23
Gold Pony Boy
152
If $C(16,r) = C(16,r+2)$, then find $r$. Explain how you know.
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
edited Dec 3 at 9:23
Tianlalu
2,9801936
asked Dec 3 at 4:23
Gold Pony Boy
152
edited Dec 3 at 9:23
Tianlalu
2,9801936
asked Dec 3 at 4:23
Gold Pony Boy
152
edited Dec 3 at 9:23
Tianlalu
2,9801936
edited Dec 3 at 9:23
Tianlalu
2,9801936
edited Dec 3 at 9:23
Tianlalu
2,9801936
2,9801936
asked Dec 3 at 4:23
Gold Pony Boy
152
asked Dec 3 at 4:23
Gold Pony Boy
152
asked Dec 3 at 4:23
Gold Pony Boy
152
152
add a comment |
add a comment |
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 be taken on at most twice 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 Dec 3 at 4:28
Display name
789313
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 at 4:52
@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
Dec 3 at 5:00
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
Dec 3 at 5:01
add a comment |
up vote
4
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 Dec 3 at 4:29
Carl Schildkraut
10.8k11439
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 at 4:45
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 at 6:06
add a comment |
up vote
2
down vote
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{r!(16-r)(15-r).(14-r)!}=frac{1}{(r+2)(r+1).r!.(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
$$implies 240-31r+r^2=r^2+3r+2$$
$$implies 34r=238$$
$$implies r=7$$
answered Dec 3 at 4:41
Fareed AF
39711
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
Dec 3 at 5:06
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 at 5:07
@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
Dec 3 at 5:12
1
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
Dec 3 at 5:30
1
This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
– smci
Dec 3 at 17:46
|
show 1 more 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 be taken on at most twice 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 Dec 3 at 4:28
Display name
789313
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 at 4:52
@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
Dec 3 at 5:00
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
Dec 3 at 5:01
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 be taken on at most twice 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 Dec 3 at 4:28
Display name
789313
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 at 4:52
@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
Dec 3 at 5:00
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
Dec 3 at 5:01
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 be taken on at most twice 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 Dec 3 at 4:28
Display name
789313
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 be taken on at most twice 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 Dec 3 at 4:28
Display name
789313
edited Dec 3 at 6:35
answered Dec 3 at 4:28
Display name
789313
answered Dec 3 at 4:28
Display name
789313
answered Dec 3 at 4:28
Display name
789313
789313
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 at 4:52
@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
Dec 3 at 5:00
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
Dec 3 at 5:01
add a comment |
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 at 4:52
@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
Dec 3 at 5:00
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
Dec 3 at 5:01
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 at 4:52
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 at 4:52
@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
Dec 3 at 5:00
@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
Dec 3 at 5:00
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
Dec 3 at 5:01
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
Dec 3 at 5:01
add a comment |
up vote
4
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 Dec 3 at 4:29
Carl Schildkraut
10.8k11439
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 at 4:45
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 at 6:06
add a comment |
up vote
4
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 Dec 3 at 4:29
Carl Schildkraut
10.8k11439
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 at 4:45
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 at 6:06
add a comment |
up vote
4
down vote
up vote
4
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 Dec 3 at 4:29
Carl Schildkraut
10.8k11439
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 Dec 3 at 4:29
Carl Schildkraut
10.8k11439
answered Dec 3 at 4:29
Carl Schildkraut
10.8k11439
answered Dec 3 at 4:29
Carl Schildkraut
10.8k11439
answered Dec 3 at 4:29
Carl Schildkraut
10.8k11439
10.8k11439
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 at 4:45
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 at 6:06
add a comment |
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 at 4:45
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 at 6:06
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 at 4:45
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 at 4:45
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 at 6:06
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 at 6:06
add a comment |
up vote
2
down vote
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{r!(16-r)(15-r).(14-r)!}=frac{1}{(r+2)(r+1).r!.(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
$$implies 240-31r+r^2=r^2+3r+2$$
$$implies 34r=238$$
$$implies r=7$$
answered Dec 3 at 4:41
Fareed AF
39711
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
Dec 3 at 5:06
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 at 5:07
@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
Dec 3 at 5:12
1
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
Dec 3 at 5:30
1
This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
– smci
Dec 3 at 17:46
|
show 1 more comment
up vote
2
down vote
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{r!(16-r)(15-r).(14-r)!}=frac{1}{(r+2)(r+1).r!.(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
$$implies 240-31r+r^2=r^2+3r+2$$
$$implies 34r=238$$
$$implies r=7$$
answered Dec 3 at 4:41
Fareed AF
39711
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
Dec 3 at 5:06
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 at 5:07
@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
Dec 3 at 5:12
1
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
Dec 3 at 5:30
1
This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
– smci
Dec 3 at 17:46
|
show 1 more comment
up vote
2
down vote
up vote
2
down vote
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{r!(16-r)(15-r).(14-r)!}=frac{1}{(r+2)(r+1).r!.(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
$$implies 240-31r+r^2=r^2+3r+2$$
$$implies 34r=238$$
$$implies r=7$$
answered Dec 3 at 4:41
Fareed AF
39711
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{r!(16-r)(15-r).(14-r)!}=frac{1}{(r+2)(r+1).r!.(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
$$implies 240-31r+r^2=r^2+3r+2$$
$$implies 34r=238$$
$$implies r=7$$
answered Dec 3 at 4:41
Fareed AF
39711
edited Dec 3 at 6:50
answered Dec 3 at 4:41
Fareed AF
39711
answered Dec 3 at 4:41
Fareed AF
39711
answered Dec 3 at 4:41
Fareed AF
39711
39711
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
Dec 3 at 5:06
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 at 5:07
@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
Dec 3 at 5:12
1
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
Dec 3 at 5:30
1
This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
– smci
Dec 3 at 17:46
|
show 1 more 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
Dec 3 at 5:06
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 at 5:07
@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
Dec 3 at 5:12
1
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
Dec 3 at 5:30
1
This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
– smci
Dec 3 at 17:46
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
Dec 3 at 5:06
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
Dec 3 at 5:06
2
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 at 5:07
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 at 5:07
@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
Dec 3 at 5:12
@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
Dec 3 at 5:12
1
1
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
Dec 3 at 5:30
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
Dec 3 at 5:30
1
1
This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
– smci
Dec 3 at 17:46
This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
– smci
Dec 3 at 17:46
|
show 1 more comment
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