sum which have binomial coefficients
$begingroup$
Finding value of
$displaystyle binom{50}{6}-binom{5}{1}binom{40}{6}+binom{5}{2}binom{30}{6}-binom{5}{3}binom{20}{6}+binom{5}{4}binom{10}{6}$
Try: Equation coefficient of $x^6$ on both side
$$bigg[(1+x)^{10}-1bigg]^5=(1+x)^{50}-binom{5}{1}(1+x)^{40}+binom{5}{2}(1+x)^{30}-binom{5}{3}(1+x)^{20}+binom{5}{4}(1+x)^{10}$$
So our required sum is coefficient of $x^6$ in $$x^5cdot bigg[1+(1+x)+(1+x)^2+cdots +(1+x)^9bigg]^5.$$
So coefficient of $x$ in $$bigg[1+(1+x)+(1+x)^2+cdots cdots +(1+x)^{9}bigg]^5$$
$$=5(10)^4 (1+2+cdots +9)=5cdot (10)^4cdot 45$$
But I am also trying to solve it using the principle of inclusion -exclusion.
I have seems that we will form $5$ groups and selecting some persons from each group, but I did not understand how I can solve it
Could some help me to solve it , thanks
combinatorics binomial-coefficients
$endgroup$
add a comment |
$begingroup$
Finding value of
$displaystyle binom{50}{6}-binom{5}{1}binom{40}{6}+binom{5}{2}binom{30}{6}-binom{5}{3}binom{20}{6}+binom{5}{4}binom{10}{6}$
Try: Equation coefficient of $x^6$ on both side
$$bigg[(1+x)^{10}-1bigg]^5=(1+x)^{50}-binom{5}{1}(1+x)^{40}+binom{5}{2}(1+x)^{30}-binom{5}{3}(1+x)^{20}+binom{5}{4}(1+x)^{10}$$
So our required sum is coefficient of $x^6$ in $$x^5cdot bigg[1+(1+x)+(1+x)^2+cdots +(1+x)^9bigg]^5.$$
So coefficient of $x$ in $$bigg[1+(1+x)+(1+x)^2+cdots cdots +(1+x)^{9}bigg]^5$$
$$=5(10)^4 (1+2+cdots +9)=5cdot (10)^4cdot 45$$
But I am also trying to solve it using the principle of inclusion -exclusion.
I have seems that we will form $5$ groups and selecting some persons from each group, but I did not understand how I can solve it
Could some help me to solve it , thanks
combinatorics binomial-coefficients
$endgroup$
3
$begingroup$
No money for a calculator ? Use wolframalpha maybe?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:27
5
$begingroup$
I bealive D Tiwari want some nice combinatorial simplification of this exspresion. Don't understand why is this voted for close.
$endgroup$
– Maria Mazur
Dec 22 '18 at 13:46
$begingroup$
Yes greedoid i want combinational argument.
$endgroup$
– DXT
Dec 22 '18 at 14:26
$begingroup$
Anyway it's a finite sum. Maybe a related to it but somehow a general one will be interesting.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:36
$begingroup$
Answer - 2250000
$endgroup$
– ablmf
Dec 24 '18 at 15:19
add a comment |
$begingroup$
Finding value of
$displaystyle binom{50}{6}-binom{5}{1}binom{40}{6}+binom{5}{2}binom{30}{6}-binom{5}{3}binom{20}{6}+binom{5}{4}binom{10}{6}$
Try: Equation coefficient of $x^6$ on both side
$$bigg[(1+x)^{10}-1bigg]^5=(1+x)^{50}-binom{5}{1}(1+x)^{40}+binom{5}{2}(1+x)^{30}-binom{5}{3}(1+x)^{20}+binom{5}{4}(1+x)^{10}$$
So our required sum is coefficient of $x^6$ in $$x^5cdot bigg[1+(1+x)+(1+x)^2+cdots +(1+x)^9bigg]^5.$$
So coefficient of $x$ in $$bigg[1+(1+x)+(1+x)^2+cdots cdots +(1+x)^{9}bigg]^5$$
$$=5(10)^4 (1+2+cdots +9)=5cdot (10)^4cdot 45$$
But I am also trying to solve it using the principle of inclusion -exclusion.
I have seems that we will form $5$ groups and selecting some persons from each group, but I did not understand how I can solve it
Could some help me to solve it , thanks
combinatorics binomial-coefficients
$endgroup$
Finding value of
$displaystyle binom{50}{6}-binom{5}{1}binom{40}{6}+binom{5}{2}binom{30}{6}-binom{5}{3}binom{20}{6}+binom{5}{4}binom{10}{6}$
Try: Equation coefficient of $x^6$ on both side
$$bigg[(1+x)^{10}-1bigg]^5=(1+x)^{50}-binom{5}{1}(1+x)^{40}+binom{5}{2}(1+x)^{30}-binom{5}{3}(1+x)^{20}+binom{5}{4}(1+x)^{10}$$
So our required sum is coefficient of $x^6$ in $$x^5cdot bigg[1+(1+x)+(1+x)^2+cdots +(1+x)^9bigg]^5.$$
So coefficient of $x$ in $$bigg[1+(1+x)+(1+x)^2+cdots cdots +(1+x)^{9}bigg]^5$$
$$=5(10)^4 (1+2+cdots +9)=5cdot (10)^4cdot 45$$
But I am also trying to solve it using the principle of inclusion -exclusion.
I have seems that we will form $5$ groups and selecting some persons from each group, but I did not understand how I can solve it
Could some help me to solve it , thanks
combinatorics binomial-coefficients
combinatorics binomial-coefficients
edited Dec 25 '18 at 0:19
Namaste
1
1
asked Dec 22 '18 at 10:21
DXTDXT
5,9012733
5,9012733
3
$begingroup$
No money for a calculator ? Use wolframalpha maybe?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:27
5
$begingroup$
I bealive D Tiwari want some nice combinatorial simplification of this exspresion. Don't understand why is this voted for close.
$endgroup$
– Maria Mazur
Dec 22 '18 at 13:46
$begingroup$
Yes greedoid i want combinational argument.
$endgroup$
– DXT
Dec 22 '18 at 14:26
$begingroup$
Anyway it's a finite sum. Maybe a related to it but somehow a general one will be interesting.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:36
$begingroup$
Answer - 2250000
$endgroup$
– ablmf
Dec 24 '18 at 15:19
add a comment |
3
$begingroup$
No money for a calculator ? Use wolframalpha maybe?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:27
5
$begingroup$
I bealive D Tiwari want some nice combinatorial simplification of this exspresion. Don't understand why is this voted for close.
$endgroup$
– Maria Mazur
Dec 22 '18 at 13:46
$begingroup$
Yes greedoid i want combinational argument.
$endgroup$
– DXT
Dec 22 '18 at 14:26
$begingroup$
Anyway it's a finite sum. Maybe a related to it but somehow a general one will be interesting.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:36
$begingroup$
Answer - 2250000
$endgroup$
– ablmf
Dec 24 '18 at 15:19
3
3
$begingroup$
No money for a calculator ? Use wolframalpha maybe?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:27
$begingroup$
No money for a calculator ? Use wolframalpha maybe?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:27
5
5
$begingroup$
I bealive D Tiwari want some nice combinatorial simplification of this exspresion. Don't understand why is this voted for close.
$endgroup$
– Maria Mazur
Dec 22 '18 at 13:46
$begingroup$
I bealive D Tiwari want some nice combinatorial simplification of this exspresion. Don't understand why is this voted for close.
$endgroup$
– Maria Mazur
Dec 22 '18 at 13:46
$begingroup$
Yes greedoid i want combinational argument.
$endgroup$
– DXT
Dec 22 '18 at 14:26
$begingroup$
Yes greedoid i want combinational argument.
$endgroup$
– DXT
Dec 22 '18 at 14:26
$begingroup$
Anyway it's a finite sum. Maybe a related to it but somehow a general one will be interesting.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:36
$begingroup$
Anyway it's a finite sum. Maybe a related to it but somehow a general one will be interesting.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:36
$begingroup$
Answer - 2250000
$endgroup$
– ablmf
Dec 24 '18 at 15:19
$begingroup$
Answer - 2250000
$endgroup$
– ablmf
Dec 24 '18 at 15:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's consider the following problem: Given 50 balls colored in five colors, 10 per color, how many ways can we choose 6 balls so that each color is represented?
Denote the answer $X$.
We will solve this problem in two different ways. Both should give the same result.
First, let's solve the above using inclusion-exclusion principle.
- We count how many ways we can choose 6 balls. That is $50 choose 6$
- Now we remove all combinations that miss a color. How many are there? There are 5 possibilities for the missing color, and 40 balls colored differently. Thus, subtract ${5 choose 1}{40 choose 6}$ combinations.
- But arrangements missing two colors were counted twice in step 2, so we need to add them back once. There are $5 choose 2$ ways to select two missing colors. This leaves 20 balls, so we have choose 6$ ways to choose the balls.
- To complete the inclusion and exclusion, we have to subtract arrangements missing 3 colors, and add arrangements missing 4 colors (there are no arrangements missing 5 colors).
In short, from inclusion-exclusion principle, we have
$$X = binom{50}{6}-binom{5}{1}binom{40}{6}+binom{5}{2}binom{30}{6}-binom{5}{3}binom{20}{6}+binom{5}{4}binom{10}{6}$$
There is, however, an easier way to solve this problem. If we select 6 balls so that there is at least one from each color, we'll have exactly one ball per color, except for one color which will have two.
So we have 5 ways to choose the color that has two balls, $10 choose 2$ ways to choose those 2 balls, and then 10 ways per color to choose the balls from the remaining colors. Thus,
$$X = 5 times {10 choose 2} times 10^4 = 5 times 45 times 10 000 = 2 250 000$$.
Thus, the sum we need to compute is 2 250 000.
$endgroup$
add a comment |
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$begingroup$
Let's consider the following problem: Given 50 balls colored in five colors, 10 per color, how many ways can we choose 6 balls so that each color is represented?
Denote the answer $X$.
We will solve this problem in two different ways. Both should give the same result.
First, let's solve the above using inclusion-exclusion principle.
- We count how many ways we can choose 6 balls. That is $50 choose 6$
- Now we remove all combinations that miss a color. How many are there? There are 5 possibilities for the missing color, and 40 balls colored differently. Thus, subtract ${5 choose 1}{40 choose 6}$ combinations.
- But arrangements missing two colors were counted twice in step 2, so we need to add them back once. There are $5 choose 2$ ways to select two missing colors. This leaves 20 balls, so we have choose 6$ ways to choose the balls.
- To complete the inclusion and exclusion, we have to subtract arrangements missing 3 colors, and add arrangements missing 4 colors (there are no arrangements missing 5 colors).
In short, from inclusion-exclusion principle, we have
$$X = binom{50}{6}-binom{5}{1}binom{40}{6}+binom{5}{2}binom{30}{6}-binom{5}{3}binom{20}{6}+binom{5}{4}binom{10}{6}$$
There is, however, an easier way to solve this problem. If we select 6 balls so that there is at least one from each color, we'll have exactly one ball per color, except for one color which will have two.
So we have 5 ways to choose the color that has two balls, $10 choose 2$ ways to choose those 2 balls, and then 10 ways per color to choose the balls from the remaining colors. Thus,
$$X = 5 times {10 choose 2} times 10^4 = 5 times 45 times 10 000 = 2 250 000$$.
Thus, the sum we need to compute is 2 250 000.
$endgroup$
add a comment |
$begingroup$
Let's consider the following problem: Given 50 balls colored in five colors, 10 per color, how many ways can we choose 6 balls so that each color is represented?
Denote the answer $X$.
We will solve this problem in two different ways. Both should give the same result.
First, let's solve the above using inclusion-exclusion principle.
- We count how many ways we can choose 6 balls. That is $50 choose 6$
- Now we remove all combinations that miss a color. How many are there? There are 5 possibilities for the missing color, and 40 balls colored differently. Thus, subtract ${5 choose 1}{40 choose 6}$ combinations.
- But arrangements missing two colors were counted twice in step 2, so we need to add them back once. There are $5 choose 2$ ways to select two missing colors. This leaves 20 balls, so we have choose 6$ ways to choose the balls.
- To complete the inclusion and exclusion, we have to subtract arrangements missing 3 colors, and add arrangements missing 4 colors (there are no arrangements missing 5 colors).
In short, from inclusion-exclusion principle, we have
$$X = binom{50}{6}-binom{5}{1}binom{40}{6}+binom{5}{2}binom{30}{6}-binom{5}{3}binom{20}{6}+binom{5}{4}binom{10}{6}$$
There is, however, an easier way to solve this problem. If we select 6 balls so that there is at least one from each color, we'll have exactly one ball per color, except for one color which will have two.
So we have 5 ways to choose the color that has two balls, $10 choose 2$ ways to choose those 2 balls, and then 10 ways per color to choose the balls from the remaining colors. Thus,
$$X = 5 times {10 choose 2} times 10^4 = 5 times 45 times 10 000 = 2 250 000$$.
Thus, the sum we need to compute is 2 250 000.
$endgroup$
add a comment |
$begingroup$
Let's consider the following problem: Given 50 balls colored in five colors, 10 per color, how many ways can we choose 6 balls so that each color is represented?
Denote the answer $X$.
We will solve this problem in two different ways. Both should give the same result.
First, let's solve the above using inclusion-exclusion principle.
- We count how many ways we can choose 6 balls. That is $50 choose 6$
- Now we remove all combinations that miss a color. How many are there? There are 5 possibilities for the missing color, and 40 balls colored differently. Thus, subtract ${5 choose 1}{40 choose 6}$ combinations.
- But arrangements missing two colors were counted twice in step 2, so we need to add them back once. There are $5 choose 2$ ways to select two missing colors. This leaves 20 balls, so we have choose 6$ ways to choose the balls.
- To complete the inclusion and exclusion, we have to subtract arrangements missing 3 colors, and add arrangements missing 4 colors (there are no arrangements missing 5 colors).
In short, from inclusion-exclusion principle, we have
$$X = binom{50}{6}-binom{5}{1}binom{40}{6}+binom{5}{2}binom{30}{6}-binom{5}{3}binom{20}{6}+binom{5}{4}binom{10}{6}$$
There is, however, an easier way to solve this problem. If we select 6 balls so that there is at least one from each color, we'll have exactly one ball per color, except for one color which will have two.
So we have 5 ways to choose the color that has two balls, $10 choose 2$ ways to choose those 2 balls, and then 10 ways per color to choose the balls from the remaining colors. Thus,
$$X = 5 times {10 choose 2} times 10^4 = 5 times 45 times 10 000 = 2 250 000$$.
Thus, the sum we need to compute is 2 250 000.
$endgroup$
Let's consider the following problem: Given 50 balls colored in five colors, 10 per color, how many ways can we choose 6 balls so that each color is represented?
Denote the answer $X$.
We will solve this problem in two different ways. Both should give the same result.
First, let's solve the above using inclusion-exclusion principle.
- We count how many ways we can choose 6 balls. That is $50 choose 6$
- Now we remove all combinations that miss a color. How many are there? There are 5 possibilities for the missing color, and 40 balls colored differently. Thus, subtract ${5 choose 1}{40 choose 6}$ combinations.
- But arrangements missing two colors were counted twice in step 2, so we need to add them back once. There are $5 choose 2$ ways to select two missing colors. This leaves 20 balls, so we have choose 6$ ways to choose the balls.
- To complete the inclusion and exclusion, we have to subtract arrangements missing 3 colors, and add arrangements missing 4 colors (there are no arrangements missing 5 colors).
In short, from inclusion-exclusion principle, we have
$$X = binom{50}{6}-binom{5}{1}binom{40}{6}+binom{5}{2}binom{30}{6}-binom{5}{3}binom{20}{6}+binom{5}{4}binom{10}{6}$$
There is, however, an easier way to solve this problem. If we select 6 balls so that there is at least one from each color, we'll have exactly one ball per color, except for one color which will have two.
So we have 5 ways to choose the color that has two balls, $10 choose 2$ ways to choose those 2 balls, and then 10 ways per color to choose the balls from the remaining colors. Thus,
$$X = 5 times {10 choose 2} times 10^4 = 5 times 45 times 10 000 = 2 250 000$$.
Thus, the sum we need to compute is 2 250 000.
edited Dec 24 '18 at 21:39
answered Dec 24 '18 at 21:12
Todor MarkovTodor Markov
2,420412
2,420412
add a comment |
add a comment |
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$begingroup$
No money for a calculator ? Use wolframalpha maybe?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:27
5
$begingroup$
I bealive D Tiwari want some nice combinatorial simplification of this exspresion. Don't understand why is this voted for close.
$endgroup$
– Maria Mazur
Dec 22 '18 at 13:46
$begingroup$
Yes greedoid i want combinational argument.
$endgroup$
– DXT
Dec 22 '18 at 14:26
$begingroup$
Anyway it's a finite sum. Maybe a related to it but somehow a general one will be interesting.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:36
$begingroup$
Answer - 2250000
$endgroup$
– ablmf
Dec 24 '18 at 15:19