Probability Problem of 2i offensive-defensive players
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
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
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
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
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 |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999852%2fprobability-problem-of-2i-offensive-defensive-players%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
use that $0!=1$
– Vasya
Nov 15 at 15:49