If $C(16,r) = C(16,r+2)$, then find $r$. Explain how you know.
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 '18 at 9:23
Tianlalu
3,09621038
asked Dec 3 '18 at 4:23
Gold Pony Boy
152
add a comment |
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 '18 at 9:23
Tianlalu
3,09621038
asked Dec 3 '18 at 4:23
Gold Pony Boy
152
add a comment |
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 '18 at 9:23
Tianlalu
3,09621038
asked Dec 3 '18 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 '18 at 9:23
Tianlalu
3,09621038
asked Dec 3 '18 at 4:23
Gold Pony Boy
152
edited Dec 3 '18 at 9:23
Tianlalu
3,09621038
asked Dec 3 '18 at 4:23
Gold Pony Boy
152
edited Dec 3 '18 at 9:23
Tianlalu
3,09621038
edited Dec 3 '18 at 9:23
Tianlalu
3,09621038
edited Dec 3 '18 at 9:23
Tianlalu
3,09621038
3,09621038
asked Dec 3 '18 at 4:23
Gold Pony Boy
152
asked Dec 3 '18 at 4:23
Gold Pony Boy
152
asked Dec 3 '18 at 4:23
Gold Pony Boy
152
152
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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 '18 at 4:28
Display name
791313
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 '18 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 '18 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 '18 at 5:01
add a comment |
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 '18 at 4:29
Carl Schildkraut
11.2k11441
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 '18 at 4:45
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 '18 at 6:06
add a comment |
$C_r^{16}=C_{r+2}^{16}$
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 '18 at 4:41
Fareed AF
44211
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 '18 at 5:06
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 '18 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 '18 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 '18 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 '18 at 17:46
|
show 1 more comment
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 '18 at 4:28
Display name
791313
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 '18 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 '18 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 '18 at 5:01
add a comment |
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 '18 at 4:28
Display name
791313
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 '18 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 '18 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 '18 at 5:01
add a comment |
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 '18 at 4:28
Display name
791313
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 '18 at 4:28
Display name
791313
edited Dec 3 '18 at 6:35
answered Dec 3 '18 at 4:28
Display name
791313
answered Dec 3 '18 at 4:28
Display name
791313
answered Dec 3 '18 at 4:28
Display name
791313
791313
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 '18 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 '18 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 '18 at 5:01
add a comment |
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 '18 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 '18 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 '18 at 5:01
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 '18 at 4:52
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
Dec 3 '18 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 '18 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 '18 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 '18 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 '18 at 5:01
add a comment |
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 '18 at 4:29
Carl Schildkraut
11.2k11441
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 '18 at 4:45
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 '18 at 6:06
add a comment |
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 '18 at 4:29
Carl Schildkraut
11.2k11441
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 '18 at 4:45
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 '18 at 6:06
add a comment |
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 '18 at 4:29
Carl Schildkraut
11.2k11441
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 '18 at 4:29
Carl Schildkraut
11.2k11441
answered Dec 3 '18 at 4:29
Carl Schildkraut
11.2k11441
answered Dec 3 '18 at 4:29
Carl Schildkraut
11.2k11441
answered Dec 3 '18 at 4:29
Carl Schildkraut
11.2k11441
11.2k11441
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 '18 at 4:45
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 '18 at 6:06
add a comment |
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 '18 at 4:45
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 '18 at 6:06
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 '18 at 4:45
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
Dec 3 '18 at 4:45
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 '18 at 6:06
@GoldPonyBoy Yes.
– Carl Schildkraut
Dec 3 '18 at 6:06
add a comment |
$C_r^{16}=C_{r+2}^{16}$
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 '18 at 4:41
Fareed AF
44211
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 '18 at 5:06
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 '18 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 '18 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 '18 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 '18 at 17:46
|
show 1 more comment
$C_r^{16}=C_{r+2}^{16}$
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 '18 at 4:41
Fareed AF
44211
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 '18 at 5:06
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 '18 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 '18 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 '18 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 '18 at 17:46
|
show 1 more comment
$C_r^{16}=C_{r+2}^{16}$
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 '18 at 4:41
Fareed AF
44211
$C_r^{16}=C_{r+2}^{16}$
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 '18 at 4:41
Fareed AF
44211
edited Dec 3 '18 at 6:50
answered Dec 3 '18 at 4:41
Fareed AF
44211
answered Dec 3 '18 at 4:41
Fareed AF
44211
answered Dec 3 '18 at 4:41
Fareed AF
44211
44211
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 '18 at 5:06
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 '18 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 '18 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 '18 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 '18 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 '18 at 5:06
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 5:06
2
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 '18 at 5:07
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
Dec 3 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 17:46
|
show 1 more comment
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