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