Combinations Example 2: Two Teams + Umpire (2 of 2: Separate Players)

[@joseph_mai]

Really helped me. I was just stuck on this question and the worked solution was confusing me even more, but this video really cleared things up thx!

[@sachithn4693]

@eddiewoo in case 2, it's actually 140 ways, because you can choose one umpire from 2 , which is 2C1=2.

[@RitikMaurya07]

yes, exactly. i was about to point that

[@frankt9156]

@@RitikMaurya07No because you also didn’t taking in account of the duplicate 4 vs 4 team. So you have to divide by 2. So in the end 2/2

[@RitikMaurya07]

@@frankt9156 ok now I'll have to watch the video again😭

[@randomv8393]

Thanks it really did help me! I wish i had a teacher like you! I mean i am literally loving maths and it's all possible because of you ❤

[@fssocrates]

8:00 I think you've missed an explanation. In case 2, as we took out the restrictions, we now need to half the number. But, on the other hand, we can choose umpire from 2 significant players. That doubles the ways to choose umpire from. So, it's easy to understand if we choose the umpire first in case 2. And that's 2C1. So, the calculation becomes: 2C1 * ( 8C4 * 4C4 ) / 2 And that's again 70.

[@KyleFES]

Why do I get 420 when I try solving by complement. The way to get 210 would involve multiplying the amount of ways they are on the same team by two (to account for the two different team). But I do not think that is right, because the teams should be interchangeable. Example: if we label the players 1 through 9, Team 1 as 1,2,3,4 and Team 2 as 5,6,7,8 is the same as Team 1 as 5,6,7,8 and Team 2 as 1,2,3,4. This would be because the teams arent actually named Team 1 and Team 2, but rather are defined by the players within the team.

[@Tharindu_LK]

In the second scenario shouldn't there be another 70 ways left where you choose the other player to be the umpire? Lets say 1 and 5 cannot be in the same team so we make 1 to umpire 9234 | 5678 | 1 _____ | _____ | 1 This scenario gives us 70 ways of choices where 1 is always the umpire Second round we mak 5 to be the umpire 1234 | 9678 | 5 _____ | _____ | 5 This also gives us 70 more unique ways Doesn't it?

[@Structuralmechanic]

I think so too. In the second case there are 2C1 ways of choosing an umpire.

[@ventin75]

Probably easier to find the answer through its complement

[@stuartkeating6508]

I don't get it. If one guy is fixed as the umpire, then wouldn't team A and team B then be interchangeable? So you have to divide by 2!??? help :( ??

[@patiwatkamonpet8745]

Yes, you got the right idea but you have 2 guys to be fixed as umpire, don't you? Once you fix one guy as umpire and split the rest into two teams, team A and team B are interchangeable so you have to divide by 2. This results in C(8,4)/2 ways but it is a partial result. Then you have to fix the other guy as umpire too so same result C(8,4)/2 ways. Summing the two cases, the net result is C(8,4).

[@Krishna-iy9vb]

@@patiwatkamonpet8745 Thanks a lot!

[@Chelsea2009FC]

boy this part of probability is confusing as fuuuuuuck

[@OmegaMindriel]

Also in case two shouldn't it be two times bigger? I mean not 70, but 140? Because the unpire can be chosen from two players.

[@yasinkhan4620]

lol I thought case 2 should be 35 cause the teams can be switched and it would be the same. Like he did in the previous video for part 1

[@tam7646]

Pretty sure you're both right. Partially. from 70 ways, the umpire can be either player, so times by 2. But then you could switch both teams around and it wouldn't make a difference to who's on which team, hence divide by 2.

[@carlyillingworth5329]

Agreed, I was waiting for the 'but' either one of those 2 people could be umpire

[@yamshukfung]

two teams can be switched, the umpire can be chosen from two players, so the result is 70 ways

[@RitikMaurya07]

@@yamshukfung not at all, in the two cases of case 2, both the selections will be different because in both, one of the dissenting players can become an umpire. and thus it will be a different selection. i believe case 2 should have 140 ways of selection