+3 votes
143 views
1. As due to rain, the match between the teams in ICC world cup got canceled , So lets the total team be 10, exclude semi finals and finals , consider only league match, What is the total number of matches that played between the teams such that each team has get exactly one match get canceled due to rain among them ?
2. let the solution for first one is n matches, how many ways those n matches can be conducted ?

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## 1 Answer

+2 votes
Best answer

before going to further refer basic question : https://gateoverflow.in/260206/sheldon-ross-example-5b-5c

we can clearly understood that teams are numbered !

therefore total matches = $\binom{\text{ no.of teams }}{2} = \binom{ 10 }{ 2 }$ = 45

now the question is " exactly one match of each team is canceled ", As we know that if one match canceled, two teams are affected !

Let there are n teams, if $\frac{n}{2}$ matches canceled, then there is a way that, every team exactly one match canceled !

Answer for part A :-  45 - (10/2) = 45 - 5 = 40

Part B :-

let there are 4 teams i.e., A,B,C, and D

==> total 4 teams = 6 matches, A-B, A-C, A-D, B-C, B-D, and C-D

for 4 teams if 2 matches canceled for " every team exactly one match canceled ! "

So, total matches according to the part A = 6-2 = 4.

in how many ways these 4 matches can be conducted ?

i) A-C, A-D, B-C, B-D ===> A-B and C-D matches canceled in this scenario

ii) A-B, A-D, B-C, C-D ===> A-C and B-D matches canceled in this scenario

iii) A-B, A-C, B-D, C-D ===> A-D and B-C matches canceled in this scenario

only these 3 are the possible ways... indirectly it is asking about, how many ways those matches can be canceled ?

for the original questions, let those 10 teams are A,B,C,D,E,F,G,H,I,J

what is the possible cancelation of 5 MATCHES ? A-B, C-D E-F, G-H, I-J

____  ____  ____ ____ ____ ===> for first blank $\frac{\binom{10}{1} * \binom{9}{1}}{2}$, for second blank  $\frac{\binom{8}{1} * \binom{7}{1}}{2}$,... for last i.e., 5th  blank  $\frac{\binom{2}{1} * \binom{1}{1}}{2}$

=  $\frac{\binom{10}{1} * \binom{9}{1} * \binom{8}{1} * \binom{7}{1} * \binom{6}{1} * \binom{5}{1} * \binom{4}{1} * \binom{3}{1} * \binom{2}{1} * \binom{1}{1} }{2.2.2.2.2}$ =  $\frac{10!}{2^5}$

and this possible is not ordered between the matches ==> divide by 5 !

Final answer for part B :-  $\frac{10!}{2^5 . 5 !}$ = 945

by Veteran (64.6k points)
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0
For the B part, when do we say if two ways are distinct?
0
sir, if atleast one of the match opponents changed

by the way we are not dealing with the dates
0
Yes. If you calculate it that way what will be the answer?
+1
sir, i calculated in that way only.
0
yes. $9 \times 7 \times 5 \times 3 \times 1 = 945.$

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