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## Here's the Solution to this Question

a)

The student MUST answer to the first 2 questions. These two questions are not fit to be chosen. So we have 13-2=11 questions left, and 10-2=8 questions left to be chosen.

$n=\binom{13-2}{10-2}={11! \over 8!(11-8)!}={11(10)(9) \over 1(2)(3)}=165$

b)

One question, the first or the second, has been answered. So we have 13-1=12 questions left, and we have to answer to 10-1=9 questions.

$n=\binom{2}{1}\binom{13-2}{10-1}=2\cdot{11! \over 9!(11-9)!}=2\cdot{11(10) \over 1(2)}=110$

c)

$n=\binom{5}{3}\binom{13-5}{10-3}={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}=10\cdot8=80$

d)

$n=\binom{5}{3}\binom{13-5}{10-3}+\binom{5}{4}\binom{13-5}{10-4}+\binom{5}{5}\binom{13-5}{10-5}$

$={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}+{5! \over 4!(5-4)!}\cdot{8! \over 6!(8-6)!}+{5! \over 5!(5-5)!}\cdot{8! \over 5!(8-5)!}$

$=10(8)+5(28)+1(56)=276$

a)

The student MUST answer to the first 2 questions. These two questions are not fit to be chosen. So we have 13-2=11 questions left, and 10-2=8 questions left to be chosen.

$n=\binom{13-2}{10-2}={11! \over 8!(11-8)!}={11(10)(9) \over 1(2)(3)}=165$

b)

One question, the first or the second, has been answered. So we have 13-1=12 questions left, and we have to answer to 10-1=9 questions.

$n=\binom{2}{1}\binom{13-2}{10-1}=2\cdot{11! \over 9!(11-9)!}=2\cdot{11(10) \over 1(2)}=110$

c)

$n=\binom{5}{3}\binom{13-5}{10-3}={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}=10\cdot8=80$

d)

$n=\binom{5}{3}\binom{13-5}{10-3}+\binom{5}{4}\binom{13-5}{10-4}+\binom{5}{5}\binom{13-5}{10-5}$

$={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}+{5! \over 4!(5-4)!}\cdot{8! \over 6!(8-6)!}+{5! \over 5!(5-5)!}\cdot{8! \over 5!(8-5)!}$

$=10(8)+5(28)+1(56)=276$

a)

The student MUST answer to the first 2 questions. These two questions are not fit to be chosen. So we have 13-2=11 questions left, and 10-2=8 questions left to be chosen.

$n=\binom{13-2}{10-2}={11! \over 8!(11-8)!}={11(10)(9) \over 1(2)(3)}=165$

b)

One question, the first or the second, has been answered. So we have 13-1=12 questions left, and we have to answer to 10-1=9 questions.

$n=\binom{2}{1}\binom{13-2}{10-1}=2\cdot{11! \over 9!(11-9)!}=2\cdot{11(10) \over 1(2)}=110$

c)

$n=\binom{5}{3}\binom{13-5}{10-3}={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}=10\cdot8=80$

d)

$n=\binom{5}{3}\binom{13-5}{10-3}+\binom{5}{4}\binom{13-5}{10-4}+\binom{5}{5}\binom{13-5}{10-5}$

$={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}+{5! \over 4!(5-4)!}\cdot{8! \over 6!(8-6)!}+{5! \over 5!(5-5)!}\cdot{8! \over 5!(8-5)!}$

$=10(8)+5(28)+1(56)=276$

a)

The student MUST answer to the first 2 questions. These two questions are not fit to be chosen. So we have 13-2=11 questions left, and 10-2=8 questions left to be chosen.

$n=\binom{13-2}{10-2}={11! \over 8!(11-8)!}={11(10)(9) \over 1(2)(3)}=165$

b)

One question, the first or the second, has been answered. So we have 13-1=12 questions left, and we have to answer to 10-1=9 questions.

$n=\binom{2}{1}\binom{13-2}{10-1}=2\cdot{11! \over 9!(11-9)!}=2\cdot{11(10) \over 1(2)}=110$

c)

$n=\binom{5}{3}\binom{13-5}{10-3}={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}=10\cdot8=80$

d)

$n=\binom{5}{3}\binom{13-5}{10-3}+\binom{5}{4}\binom{13-5}{10-4}+\binom{5}{5}\binom{13-5}{10-5}$

$={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}+{5! \over 4!(5-4)!}\cdot{8! \over 6!(8-6)!}+{5! \over 5!(5-5)!}\cdot{8! \over 5!(8-5)!}$

$=10(8)+5(28)+1(56)=276$

a)

The student MUST answer to the first 2 questions. These two questions are not fit to be chosen. So we have 13-2=11 questions left, and 10-2=8 questions left to be chosen.

$n=\binom{13-2}{10-2}={11! \over 8!(11-8)!}={11(10)(9) \over 1(2)(3)}=165$

b)

One question, the first or the second, has been answered. So we have 13-1=12 questions left, and we have to answer to 10-1=9 questions.

$n=\binom{2}{1}\binom{13-2}{10-1}=2\cdot{11! \over 9!(11-9)!}=2\cdot{11(10) \over 1(2)}=110$

c)

$n=\binom{5}{3}\binom{13-5}{10-3}={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}=10\cdot8=80$

d)

$n=\binom{5}{3}\binom{13-5}{10-3}+\binom{5}{4}\binom{13-5}{10-4}+\binom{5}{5}\binom{13-5}{10-5}$

$={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}+{5! \over 4!(5-4)!}\cdot{8! \over 6!(8-6)!}+{5! \over 5!(5-5)!}\cdot{8! \over 5!(8-5)!}$

$=10(8)+5(28)+1(56)=276$

a)

The student MUST answer to the first 2 questions. These two questions are not fit to be chosen. So we have 13-2=11 questions left, and 10-2=8 questions left to be chosen.

$n=\binom{13-2}{10-2}={11! \over 8!(11-8)!}={11(10)(9) \over 1(2)(3)}=165$

b)

One question, the first or the second, has been answered. So we have 13-1=12 questions left, and we have to answer to 10-1=9 questions.

$n=\binom{2}{1}\binom{13-2}{10-1}=2\cdot{11! \over 9!(11-9)!}=2\cdot{11(10) \over 1(2)}=110$

c)

$n=\binom{5}{3}\binom{13-5}{10-3}={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}=10\cdot8=80$

d)

$n=\binom{5}{3}\binom{13-5}{10-3}+\binom{5}{4}\binom{13-5}{10-4}+\binom{5}{5}\binom{13-5}{10-5}$

$={5! \over 3!(5-3)!}\cdot{8! \over 7!(8-7)!}+{5! \over 4!(5-4)!}\cdot{8! \over 6!(8-6)!}+{5! \over 5!(5-5)!}\cdot{8! \over 5!(8-5)!}$

$=10(8)+5(28)+1(56)=276$