Probability Problem of 2i offensive-defensive players
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Question: A football team consists of 20 offensive and 20 defensive players. The players are to be paired in groups of 2 for the purpose of determining roommates. What is the probability that there are 2i offensive–defensive roommate pairs, i = 1,2,...,10?
20 Offensive (O), 20 Defensive (D) ==> 40 people total and 20 pairs
The question asks for the probability of 2i OD pairs. If we have 2i OD pairs, then we have (20−2i) OO and DD pairs. So I think the solution will be of form:
$frac{(mbox{total OD pair combinations}) cdot (mbox{total OO Comb)(total DD comb})}{mbox{total unordered pair combinations} }$
Therefore, the answer will be:
$cfrac{binom{20}{2i}^2(2i)! left[ cfrac{(20-2i)!}{2^{10-i}(10-i)!} right]^2} {cfrac{40!}{2^{20}20! }} i=0,1,...,10$
My question is, when i=10, the numerator should be logically $(20!)$
But here the second part of the numerator becomes $cfrac{0}{0}$. So I can't proceed further.
Please help me to solve this problem. Thanks in advance.
probability combinatorics
asked Nov 15 at 15:42
user587389
326
add a comment |
up vote
1
down vote
favorite
Question: A football team consists of 20 offensive and 20 defensive players. The players are to be paired in groups of 2 for the purpose of determining roommates. What is the probability that there are 2i offensive–defensive roommate pairs, i = 1,2,...,10?
20 Offensive (O), 20 Defensive (D) ==> 40 people total and 20 pairs
The question asks for the probability of 2i OD pairs. If we have 2i OD pairs, then we have (20−2i) OO and DD pairs. So I think the solution will be of form:
$frac{(mbox{total OD pair combinations}) cdot (mbox{total OO Comb)(total DD comb})}{mbox{total unordered pair combinations} }$
Therefore, the answer will be:
$cfrac{binom{20}{2i}^2(2i)! left[ cfrac{(20-2i)!}{2^{10-i}(10-i)!} right]^2} {cfrac{40!}{2^{20}20! }} i=0,1,...,10$
My question is, when i=10, the numerator should be logically $(20!)$
But here the second part of the numerator becomes $cfrac{0}{0}$. So I can't proceed further.
Please help me to solve this problem. Thanks in advance.
probability combinatorics
asked Nov 15 at 15:42
user587389
326
3
use that $0!=1$
– Vasya
Nov 15 at 15:49
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Question: A football team consists of 20 offensive and 20 defensive players. The players are to be paired in groups of 2 for the purpose of determining roommates. What is the probability that there are 2i offensive–defensive roommate pairs, i = 1,2,...,10?
20 Offensive (O), 20 Defensive (D) ==> 40 people total and 20 pairs
The question asks for the probability of 2i OD pairs. If we have 2i OD pairs, then we have (20−2i) OO and DD pairs. So I think the solution will be of form:
$frac{(mbox{total OD pair combinations}) cdot (mbox{total OO Comb)(total DD comb})}{mbox{total unordered pair combinations} }$
Therefore, the answer will be:
$cfrac{binom{20}{2i}^2(2i)! left[ cfrac{(20-2i)!}{2^{10-i}(10-i)!} right]^2} {cfrac{40!}{2^{20}20! }} i=0,1,...,10$
My question is, when i=10, the numerator should be logically $(20!)$
But here the second part of the numerator becomes $cfrac{0}{0}$. So I can't proceed further.
Please help me to solve this problem. Thanks in advance.
probability combinatorics
asked Nov 15 at 15:42
user587389
326
Question: A football team consists of 20 offensive and 20 defensive players. The players are to be paired in groups of 2 for the purpose of determining roommates. What is the probability that there are 2i offensive–defensive roommate pairs, i = 1,2,...,10?
20 Offensive (O), 20 Defensive (D) ==> 40 people total and 20 pairs
The question asks for the probability of 2i OD pairs. If we have 2i OD pairs, then we have (20−2i) OO and DD pairs. So I think the solution will be of form:
$frac{(mbox{total OD pair combinations}) cdot (mbox{total OO Comb)(total DD comb})}{mbox{total unordered pair combinations} }$
Therefore, the answer will be:
$cfrac{binom{20}{2i}^2(2i)! left[ cfrac{(20-2i)!}{2^{10-i}(10-i)!} right]^2} {cfrac{40!}{2^{20}20! }} i=0,1,...,10$
My question is, when i=10, the numerator should be logically $(20!)$
But here the second part of the numerator becomes $cfrac{0}{0}$. So I can't proceed further.
Please help me to solve this problem. Thanks in advance.
probability combinatorics
probability combinatorics
asked Nov 15 at 15:42
user587389
326
asked Nov 15 at 15:42
user587389
326
asked Nov 15 at 15:42
user587389
326
asked Nov 15 at 15:42
user587389
326
asked Nov 15 at 15:42
user587389
326
326
3
use that $0!=1$
– Vasya
Nov 15 at 15:49
add a comment |
3
use that $0!=1$
– Vasya
Nov 15 at 15:49
3
3
use that $0!=1$
– Vasya
Nov 15 at 15:49
use that $0!=1$
– Vasya
Nov 15 at 15:49
add a comment |
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3
use that $0!=1$
– Vasya
Nov 15 at 15:49