Calculating number of combinations of multiple sets, each containing different number of elements
$begingroup$
I'm not a math genius so please consider that when posting your explanation.
I have the following sets, arbitrarily named:
a [a1, a2, a3]
b [b1, b2]
c [c1, c2, c3, c4]
d [d1, d2, d3, d4, d5]
e [e1, e2, e3]
f [f1, f2]
g [g1]
Question #1) What is the total number of combinations if I have to select one element from each set? I think I know this answer, but I want to confirm. I believe this is the rule of products, so the answer is the number of elements in each set multiplied together.
In the case above, it would be:
3 * 2 * 4 * 5 * 3 * 2 * 1 = 720 combinations
Question #2) What is the total number of combinations if I have to select one element from set a, one element from set b, and one element from 4 of the 5 following sets: c, d, e, f, g?
How would this answer change if there was another set 'h' with 6 elements [h1, h2, h3, h4, h5, h6] and I had to still select one element each from set a and b, but now had to select one element each from 4 of the now 6 sets (c, d, e, f, g, h)?
Thank you for the help in advance.
combinations
asked Feb 5 '16 at 13:59
mathfrogmathfrog
64
$endgroup$
add a comment |
$begingroup$
I'm not a math genius so please consider that when posting your explanation.
I have the following sets, arbitrarily named:
a [a1, a2, a3]
b [b1, b2]
c [c1, c2, c3, c4]
d [d1, d2, d3, d4, d5]
e [e1, e2, e3]
f [f1, f2]
g [g1]
Question #1) What is the total number of combinations if I have to select one element from each set? I think I know this answer, but I want to confirm. I believe this is the rule of products, so the answer is the number of elements in each set multiplied together.
In the case above, it would be:
3 * 2 * 4 * 5 * 3 * 2 * 1 = 720 combinations
Question #2) What is the total number of combinations if I have to select one element from set a, one element from set b, and one element from 4 of the 5 following sets: c, d, e, f, g?
How would this answer change if there was another set 'h' with 6 elements [h1, h2, h3, h4, h5, h6] and I had to still select one element each from set a and b, but now had to select one element each from 4 of the now 6 sets (c, d, e, f, g, h)?
Thank you for the help in advance.
combinations
asked Feb 5 '16 at 13:59
mathfrogmathfrog
64
$endgroup$
$begingroup$
Your answer to question 1 is correct. Please show us your attempt at question 2.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 14:07
$begingroup$
So in the case I listed above:
$endgroup$
– mathfrog
Feb 5 '16 at 14:08
$begingroup$
Ok, here goes for the 1st part of question #2: 3 * 2 * (5 choose 4) * [(4 choose 1) * (5 choose 1) * (3 choose 1) * (2 choose 1) * (1 choose 1)] = 3600? For 2nd part of question #2 would it be: 3 * 2 * (6 choose 4) * [(4 choose 1) * (5 choose 1) * (3 choose 1) * (2 choose 1) * (1 choose 1) * (6 choose 1)] = 64800?
$endgroup$
– mathfrog
Feb 5 '16 at 14:15
add a comment |
$begingroup$
I'm not a math genius so please consider that when posting your explanation.
I have the following sets, arbitrarily named:
a [a1, a2, a3]
b [b1, b2]
c [c1, c2, c3, c4]
d [d1, d2, d3, d4, d5]
e [e1, e2, e3]
f [f1, f2]
g [g1]
Question #1) What is the total number of combinations if I have to select one element from each set? I think I know this answer, but I want to confirm. I believe this is the rule of products, so the answer is the number of elements in each set multiplied together.
In the case above, it would be:
3 * 2 * 4 * 5 * 3 * 2 * 1 = 720 combinations
Question #2) What is the total number of combinations if I have to select one element from set a, one element from set b, and one element from 4 of the 5 following sets: c, d, e, f, g?
How would this answer change if there was another set 'h' with 6 elements [h1, h2, h3, h4, h5, h6] and I had to still select one element each from set a and b, but now had to select one element each from 4 of the now 6 sets (c, d, e, f, g, h)?
Thank you for the help in advance.
combinations
asked Feb 5 '16 at 13:59
mathfrogmathfrog
64
$endgroup$
I'm not a math genius so please consider that when posting your explanation.
I have the following sets, arbitrarily named:
a [a1, a2, a3]
b [b1, b2]
c [c1, c2, c3, c4]
d [d1, d2, d3, d4, d5]
e [e1, e2, e3]
f [f1, f2]
g [g1]
Question #1) What is the total number of combinations if I have to select one element from each set? I think I know this answer, but I want to confirm. I believe this is the rule of products, so the answer is the number of elements in each set multiplied together.
In the case above, it would be:
3 * 2 * 4 * 5 * 3 * 2 * 1 = 720 combinations
Question #2) What is the total number of combinations if I have to select one element from set a, one element from set b, and one element from 4 of the 5 following sets: c, d, e, f, g?
How would this answer change if there was another set 'h' with 6 elements [h1, h2, h3, h4, h5, h6] and I had to still select one element each from set a and b, but now had to select one element each from 4 of the now 6 sets (c, d, e, f, g, h)?
Thank you for the help in advance.
combinations
combinations
asked Feb 5 '16 at 13:59
mathfrogmathfrog
64
asked Feb 5 '16 at 13:59
mathfrogmathfrog
64
edited Feb 5 '16 at 13:59
mathfrog
asked Feb 5 '16 at 13:59
mathfrogmathfrog
64
asked Feb 5 '16 at 13:59
mathfrogmathfrog
64
asked Feb 5 '16 at 13:59
mathfrogmathfrog
64
64
$begingroup$
Your answer to question 1 is correct. Please show us your attempt at question 2.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 14:07
$begingroup$
So in the case I listed above:
$endgroup$
– mathfrog
Feb 5 '16 at 14:08
$begingroup$
Ok, here goes for the 1st part of question #2: 3 * 2 * (5 choose 4) * [(4 choose 1) * (5 choose 1) * (3 choose 1) * (2 choose 1) * (1 choose 1)] = 3600? For 2nd part of question #2 would it be: 3 * 2 * (6 choose 4) * [(4 choose 1) * (5 choose 1) * (3 choose 1) * (2 choose 1) * (1 choose 1) * (6 choose 1)] = 64800?
$endgroup$
– mathfrog
Feb 5 '16 at 14:15
add a comment |
$begingroup$
Your answer to question 1 is correct. Please show us your attempt at question 2.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 14:07
$begingroup$
So in the case I listed above:
$endgroup$
– mathfrog
Feb 5 '16 at 14:08
$begingroup$
Ok, here goes for the 1st part of question #2: 3 * 2 * (5 choose 4) * [(4 choose 1) * (5 choose 1) * (3 choose 1) * (2 choose 1) * (1 choose 1)] = 3600? For 2nd part of question #2 would it be: 3 * 2 * (6 choose 4) * [(4 choose 1) * (5 choose 1) * (3 choose 1) * (2 choose 1) * (1 choose 1) * (6 choose 1)] = 64800?
$endgroup$
– mathfrog
Feb 5 '16 at 14:15
$begingroup$
Your answer to question 1 is correct. Please show us your attempt at question 2.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 14:07
$begingroup$
Your answer to question 1 is correct. Please show us your attempt at question 2.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 14:07
$begingroup$
So in the case I listed above:
$endgroup$
– mathfrog
Feb 5 '16 at 14:08
$begingroup$
So in the case I listed above:
$endgroup$
– mathfrog
Feb 5 '16 at 14:08
$begingroup$
Ok, here goes for the 1st part of question #2: 3 * 2 * (5 choose 4) * [(4 choose 1) * (5 choose 1) * (3 choose 1) * (2 choose 1) * (1 choose 1)] = 3600? For 2nd part of question #2 would it be: 3 * 2 * (6 choose 4) * [(4 choose 1) * (5 choose 1) * (3 choose 1) * (2 choose 1) * (1 choose 1) * (6 choose 1)] = 64800?
$endgroup$
– mathfrog
Feb 5 '16 at 14:15
$begingroup$
Ok, here goes for the 1st part of question #2: 3 * 2 * (5 choose 4) * [(4 choose 1) * (5 choose 1) * (3 choose 1) * (2 choose 1) * (1 choose 1)] = 3600? For 2nd part of question #2 would it be: 3 * 2 * (6 choose 4) * [(4 choose 1) * (5 choose 1) * (3 choose 1) * (2 choose 1) * (1 choose 1) * (6 choose 1)] = 64800?
$endgroup$
– mathfrog
Feb 5 '16 at 14:15
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Your answer to the first question is correct.
You can apply the same method to the second question.
Selections from sets a, b, c, d, e, f: $3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 2$
Selections from sets a, b, c, d, e, g: $3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 1$
Selections from sets a, b, c, d, f, g: $3 cdot 2 cdot 4 cdot 5 cdot 2 cdot 1$
Selections from sets a, b, c, e, f, g: $3 cdot 2 cdot 4 cdot 3 cdot 2 cdot 1$
Selections from sets a, b, d, e, f, g: $3 cdot 2 cdot 5 cdot 3 cdot 2 cdot 1$
To find the number of selections from sets a and b and four of the five sets c, d, e, f, g, we add the above results to obtain
begin{align*}
3 cdot 2 cdot & (4 cdot 5 cdot 3 cdot 2 + 4 cdot 5 cdot 3 cdot 1 + 4 cdot 5 cdot 2 cdot 1 + 4 cdot 3 cdot 2 cdot 1 + 5 cdot3 cdot 2 cdot 1)\
& = 3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 2 cdot 1left(frac{1}{1} + frac{1}{2} + frac{1}{3} + frac{1}{5} + frac{1}{4}right)
end{align*}
The same method could be applied to your third question, but since there are
$binom{6}{4} = 15$ cases, calculating the answer is tedious. The cases are selections from the sets:
a,b,c,d,e,f
a,b,c,d,e,g
a,b,c,d,e,h
a,b,c,d,f,g
a,b,c,d,f,h
a,b,c,d,g,h
a,b,c,e,f,g
a,b,c,e,f,h
a,b,c,e,g,h
a,b,c,f,g,h
a,b,d,e,f,g
a,b,d,e,f,h
a,b,d,e,g,h
a,b,d,f,g,h
a,b,e,f,g,h
answered Feb 5 '16 at 14:31
N. F. TaussigN. F. Taussig
44.1k93356
$endgroup$
$begingroup$
Thank you for that, is there a way to express the above 2 situations you answered with an equation? I'm trying to help some software engineers make computer code that will calculate various situations of multiple sets with differing number of elements. It will always be the situation where one element must be chosen from sets a and b, and one element will be chosen from _ of _ sets c through z.
$endgroup$
– mathfrog
Feb 5 '16 at 15:15
$begingroup$
Since there are different numbers of elements in the sets, this problem does not lend itself to a nice answer. I just used the Multiplication Principle and the Addition Principle.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 15:23
$begingroup$
@mathfrog I revised my answer to the first part of the second question. I do not know enough about programming to know whether you can generalize it efficiently.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 16:07
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
votes
$begingroup$
Your answer to the first question is correct.
You can apply the same method to the second question.
Selections from sets a, b, c, d, e, f: $3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 2$
Selections from sets a, b, c, d, e, g: $3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 1$
Selections from sets a, b, c, d, f, g: $3 cdot 2 cdot 4 cdot 5 cdot 2 cdot 1$
Selections from sets a, b, c, e, f, g: $3 cdot 2 cdot 4 cdot 3 cdot 2 cdot 1$
Selections from sets a, b, d, e, f, g: $3 cdot 2 cdot 5 cdot 3 cdot 2 cdot 1$
To find the number of selections from sets a and b and four of the five sets c, d, e, f, g, we add the above results to obtain
begin{align*}
3 cdot 2 cdot & (4 cdot 5 cdot 3 cdot 2 + 4 cdot 5 cdot 3 cdot 1 + 4 cdot 5 cdot 2 cdot 1 + 4 cdot 3 cdot 2 cdot 1 + 5 cdot3 cdot 2 cdot 1)\
& = 3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 2 cdot 1left(frac{1}{1} + frac{1}{2} + frac{1}{3} + frac{1}{5} + frac{1}{4}right)
end{align*}
The same method could be applied to your third question, but since there are
$binom{6}{4} = 15$ cases, calculating the answer is tedious. The cases are selections from the sets:
a,b,c,d,e,f
a,b,c,d,e,g
a,b,c,d,e,h
a,b,c,d,f,g
a,b,c,d,f,h
a,b,c,d,g,h
a,b,c,e,f,g
a,b,c,e,f,h
a,b,c,e,g,h
a,b,c,f,g,h
a,b,d,e,f,g
a,b,d,e,f,h
a,b,d,e,g,h
a,b,d,f,g,h
a,b,e,f,g,h
answered Feb 5 '16 at 14:31
N. F. TaussigN. F. Taussig
44.1k93356
$endgroup$
$begingroup$
Thank you for that, is there a way to express the above 2 situations you answered with an equation? I'm trying to help some software engineers make computer code that will calculate various situations of multiple sets with differing number of elements. It will always be the situation where one element must be chosen from sets a and b, and one element will be chosen from _ of _ sets c through z.
$endgroup$
– mathfrog
Feb 5 '16 at 15:15
$begingroup$
Since there are different numbers of elements in the sets, this problem does not lend itself to a nice answer. I just used the Multiplication Principle and the Addition Principle.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 15:23
$begingroup$
@mathfrog I revised my answer to the first part of the second question. I do not know enough about programming to know whether you can generalize it efficiently.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 16:07
add a comment |
$begingroup$
Your answer to the first question is correct.
You can apply the same method to the second question.
Selections from sets a, b, c, d, e, f: $3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 2$
Selections from sets a, b, c, d, e, g: $3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 1$
Selections from sets a, b, c, d, f, g: $3 cdot 2 cdot 4 cdot 5 cdot 2 cdot 1$
Selections from sets a, b, c, e, f, g: $3 cdot 2 cdot 4 cdot 3 cdot 2 cdot 1$
Selections from sets a, b, d, e, f, g: $3 cdot 2 cdot 5 cdot 3 cdot 2 cdot 1$
To find the number of selections from sets a and b and four of the five sets c, d, e, f, g, we add the above results to obtain
begin{align*}
3 cdot 2 cdot & (4 cdot 5 cdot 3 cdot 2 + 4 cdot 5 cdot 3 cdot 1 + 4 cdot 5 cdot 2 cdot 1 + 4 cdot 3 cdot 2 cdot 1 + 5 cdot3 cdot 2 cdot 1)\
& = 3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 2 cdot 1left(frac{1}{1} + frac{1}{2} + frac{1}{3} + frac{1}{5} + frac{1}{4}right)
end{align*}
The same method could be applied to your third question, but since there are
$binom{6}{4} = 15$ cases, calculating the answer is tedious. The cases are selections from the sets:
a,b,c,d,e,f
a,b,c,d,e,g
a,b,c,d,e,h
a,b,c,d,f,g
a,b,c,d,f,h
a,b,c,d,g,h
a,b,c,e,f,g
a,b,c,e,f,h
a,b,c,e,g,h
a,b,c,f,g,h
a,b,d,e,f,g
a,b,d,e,f,h
a,b,d,e,g,h
a,b,d,f,g,h
a,b,e,f,g,h
answered Feb 5 '16 at 14:31
N. F. TaussigN. F. Taussig
44.1k93356
$endgroup$
$begingroup$
Thank you for that, is there a way to express the above 2 situations you answered with an equation? I'm trying to help some software engineers make computer code that will calculate various situations of multiple sets with differing number of elements. It will always be the situation where one element must be chosen from sets a and b, and one element will be chosen from _ of _ sets c through z.
$endgroup$
– mathfrog
Feb 5 '16 at 15:15
$begingroup$
Since there are different numbers of elements in the sets, this problem does not lend itself to a nice answer. I just used the Multiplication Principle and the Addition Principle.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 15:23
$begingroup$
@mathfrog I revised my answer to the first part of the second question. I do not know enough about programming to know whether you can generalize it efficiently.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 16:07
add a comment |
$begingroup$
Your answer to the first question is correct.
You can apply the same method to the second question.
Selections from sets a, b, c, d, e, f: $3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 2$
Selections from sets a, b, c, d, e, g: $3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 1$
Selections from sets a, b, c, d, f, g: $3 cdot 2 cdot 4 cdot 5 cdot 2 cdot 1$
Selections from sets a, b, c, e, f, g: $3 cdot 2 cdot 4 cdot 3 cdot 2 cdot 1$
Selections from sets a, b, d, e, f, g: $3 cdot 2 cdot 5 cdot 3 cdot 2 cdot 1$
To find the number of selections from sets a and b and four of the five sets c, d, e, f, g, we add the above results to obtain
begin{align*}
3 cdot 2 cdot & (4 cdot 5 cdot 3 cdot 2 + 4 cdot 5 cdot 3 cdot 1 + 4 cdot 5 cdot 2 cdot 1 + 4 cdot 3 cdot 2 cdot 1 + 5 cdot3 cdot 2 cdot 1)\
& = 3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 2 cdot 1left(frac{1}{1} + frac{1}{2} + frac{1}{3} + frac{1}{5} + frac{1}{4}right)
end{align*}
The same method could be applied to your third question, but since there are
$binom{6}{4} = 15$ cases, calculating the answer is tedious. The cases are selections from the sets:
a,b,c,d,e,f
a,b,c,d,e,g
a,b,c,d,e,h
a,b,c,d,f,g
a,b,c,d,f,h
a,b,c,d,g,h
a,b,c,e,f,g
a,b,c,e,f,h
a,b,c,e,g,h
a,b,c,f,g,h
a,b,d,e,f,g
a,b,d,e,f,h
a,b,d,e,g,h
a,b,d,f,g,h
a,b,e,f,g,h
answered Feb 5 '16 at 14:31
N. F. TaussigN. F. Taussig
44.1k93356
$endgroup$
Your answer to the first question is correct.
You can apply the same method to the second question.
Selections from sets a, b, c, d, e, f: $3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 2$
Selections from sets a, b, c, d, e, g: $3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 1$
Selections from sets a, b, c, d, f, g: $3 cdot 2 cdot 4 cdot 5 cdot 2 cdot 1$
Selections from sets a, b, c, e, f, g: $3 cdot 2 cdot 4 cdot 3 cdot 2 cdot 1$
Selections from sets a, b, d, e, f, g: $3 cdot 2 cdot 5 cdot 3 cdot 2 cdot 1$
To find the number of selections from sets a and b and four of the five sets c, d, e, f, g, we add the above results to obtain
begin{align*}
3 cdot 2 cdot & (4 cdot 5 cdot 3 cdot 2 + 4 cdot 5 cdot 3 cdot 1 + 4 cdot 5 cdot 2 cdot 1 + 4 cdot 3 cdot 2 cdot 1 + 5 cdot3 cdot 2 cdot 1)\
& = 3 cdot 2 cdot 4 cdot 5 cdot 3 cdot 2 cdot 1left(frac{1}{1} + frac{1}{2} + frac{1}{3} + frac{1}{5} + frac{1}{4}right)
end{align*}
The same method could be applied to your third question, but since there are
$binom{6}{4} = 15$ cases, calculating the answer is tedious. The cases are selections from the sets:
a,b,c,d,e,f
a,b,c,d,e,g
a,b,c,d,e,h
a,b,c,d,f,g
a,b,c,d,f,h
a,b,c,d,g,h
a,b,c,e,f,g
a,b,c,e,f,h
a,b,c,e,g,h
a,b,c,f,g,h
a,b,d,e,f,g
a,b,d,e,f,h
a,b,d,e,g,h
a,b,d,f,g,h
a,b,e,f,g,h
answered Feb 5 '16 at 14:31
N. F. TaussigN. F. Taussig
44.1k93356
edited Feb 5 '16 at 16:07
answered Feb 5 '16 at 14:31
N. F. TaussigN. F. Taussig
44.1k93356
answered Feb 5 '16 at 14:31
N. F. TaussigN. F. Taussig
44.1k93356
answered Feb 5 '16 at 14:31
N. F. TaussigN. F. Taussig
44.1k93356
44.1k93356
$begingroup$
Thank you for that, is there a way to express the above 2 situations you answered with an equation? I'm trying to help some software engineers make computer code that will calculate various situations of multiple sets with differing number of elements. It will always be the situation where one element must be chosen from sets a and b, and one element will be chosen from _ of _ sets c through z.
$endgroup$
– mathfrog
Feb 5 '16 at 15:15
$begingroup$
Since there are different numbers of elements in the sets, this problem does not lend itself to a nice answer. I just used the Multiplication Principle and the Addition Principle.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 15:23
$begingroup$
@mathfrog I revised my answer to the first part of the second question. I do not know enough about programming to know whether you can generalize it efficiently.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 16:07
add a comment |
$begingroup$
Thank you for that, is there a way to express the above 2 situations you answered with an equation? I'm trying to help some software engineers make computer code that will calculate various situations of multiple sets with differing number of elements. It will always be the situation where one element must be chosen from sets a and b, and one element will be chosen from _ of _ sets c through z.
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– mathfrog
Feb 5 '16 at 15:15
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Since there are different numbers of elements in the sets, this problem does not lend itself to a nice answer. I just used the Multiplication Principle and the Addition Principle.
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– N. F. Taussig
Feb 5 '16 at 15:23
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@mathfrog I revised my answer to the first part of the second question. I do not know enough about programming to know whether you can generalize it efficiently.
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– N. F. Taussig
Feb 5 '16 at 16:07
$begingroup$
Thank you for that, is there a way to express the above 2 situations you answered with an equation? I'm trying to help some software engineers make computer code that will calculate various situations of multiple sets with differing number of elements. It will always be the situation where one element must be chosen from sets a and b, and one element will be chosen from _ of _ sets c through z.
$endgroup$
– mathfrog
Feb 5 '16 at 15:15
$begingroup$
Thank you for that, is there a way to express the above 2 situations you answered with an equation? I'm trying to help some software engineers make computer code that will calculate various situations of multiple sets with differing number of elements. It will always be the situation where one element must be chosen from sets a and b, and one element will be chosen from _ of _ sets c through z.
$endgroup$
– mathfrog
Feb 5 '16 at 15:15
$begingroup$
Since there are different numbers of elements in the sets, this problem does not lend itself to a nice answer. I just used the Multiplication Principle and the Addition Principle.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 15:23
$begingroup$
Since there are different numbers of elements in the sets, this problem does not lend itself to a nice answer. I just used the Multiplication Principle and the Addition Principle.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 15:23
$begingroup$
@mathfrog I revised my answer to the first part of the second question. I do not know enough about programming to know whether you can generalize it efficiently.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 16:07
$begingroup$
@mathfrog I revised my answer to the first part of the second question. I do not know enough about programming to know whether you can generalize it efficiently.
$endgroup$
– N. F. Taussig
Feb 5 '16 at 16:07
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$begingroup$
Your answer to question 1 is correct. Please show us your attempt at question 2.
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– N. F. Taussig
Feb 5 '16 at 14:07
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So in the case I listed above:
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– mathfrog
Feb 5 '16 at 14:08
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Ok, here goes for the 1st part of question #2: 3 * 2 * (5 choose 4) * [(4 choose 1) * (5 choose 1) * (3 choose 1) * (2 choose 1) * (1 choose 1)] = 3600? For 2nd part of question #2 would it be: 3 * 2 * (6 choose 4) * [(4 choose 1) * (5 choose 1) * (3 choose 1) * (2 choose 1) * (1 choose 1) * (6 choose 1)] = 64800?
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– mathfrog
Feb 5 '16 at 14:15