A student

A student is to answer 8 out of 10 questions on the exam.
a) find the number n of ways the student can choose 8 out of 10 questions
b) find n if the student must answer the first three questions
c) How many if he must answer at least 4 of the first 5 questions?

Result

a =  45
b =  21
c =  35

Solution:

C8(10)=(108)=10!8!(10−8)!=10⋅92⋅1=45  a=(810)=45C_{{ 8}}(10) = \dbinom{ 10}{ 8} = \dfrac{ 10! }{ 8!(10-8)!} = \dfrac{ 10 \cdot 9 } { 2 \cdot 1 } = 45 \ \\ \ \\ a = { { 8 } \choose 10 } = 45
C5(7)=(75)=7!5!(7−5)!=7⋅62⋅1=21  b=(8−310−3)=21C_{{ 5}}(7) = \dbinom{ 7}{ 5} = \dfrac{ 7! }{ 5!(7-5)!} = \dfrac{ 7 \cdot 6 } { 2 \cdot 1 } = 21 \ \\ \ \\ b = { { 8-3 } \choose 10-3 } = 21
C5(5)=(55)=5!5!(5−5)!=11=1  C3(5)=(53)=5!3!(5−3)!=5⋅42⋅1=10  c=(45)⋅ (45)+(55)⋅ (53)=(45)⋅ (45)+1⋅ 10=35C4(5)=(54)=5!4!(5−4)!=51=5  C5(5)=(55)=5!5!(5−5)!=11=1  C3(5)=(53)=5!3!(5−3)!=5⋅42⋅1=10  c=(45)⋅ (45)+(55)⋅ (53)=(45)⋅ (45)+1⋅ 10=35C_{{ 5}}(5) = \dbinom{ 5}{ 5} = \dfrac{ 5! }{ 5!(5-5)!} = \dfrac{ 1 } { 1 } = 1 \ \\ \ \\ C_{{ 3}}(5) = \dbinom{ 5}{ 3} = \dfrac{ 5! }{ 3!(5-3)!} = \dfrac{ 5 \cdot 4 } { 2 \cdot 1 } = 10 \ \\ \ \\ c = { { 4 } \choose 5 } \cdot \ { { 4 } \choose 5 } + { { 5 } \choose 5 } \cdot \ { { 5 } \choose 3 } = { { 4 } \choose 5 } \cdot \ { { 4 } \choose 5 } + 1 \cdot \ 10 = 35C_{{ 4}}(5) = \dbinom{ 5}{ 4} = \dfrac{ 5! }{ 4!(5-4)!} = \dfrac{ 5 } { 1 } = 5 \ \\ \ \\ C_{{ 5}}(5) = \dbinom{ 5}{ 5} = \dfrac{ 5! }{ 5!(5-5)!} = \dfrac{ 1 } { 1 } = 1 \ \\ \ \\ C_{{ 3}}(5) = \dbinom{ 5}{ 3} = \dfrac{ 5! }{ 3!(5-3)!} = \dfrac{ 5 \cdot 4 } { 2 \cdot 1 } = 10 \ \\ \ \\ c = { { 4 } \choose 5 } \cdot \ { { 4 } \choose 5 } + { { 5 } \choose 5 } \cdot \ { { 5 } \choose 3 } = { { 4 } \choose 5 } \cdot \ { { 4 } \choose 5 } + 1 \cdot \ 10 = 35

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