Question

6) Calcule:
a) A 5,2
b) A n,2 = 30

7) Resolva:
a) C 7,5
b) C n,4 = 4 C n,3

8) Determine o valor de n da equação:
( n+1)! = 12
( n -1)!

Answer 1

Resposta:

Explicação passo-a-passo:

A n,p = n!/(n-p)!

6)

A)

A 5,2 = 5!/(5-2)! = 5!/3! = 5.4.3!/3! = 20

R.: 20

------------

B)

A n,2 = 30

n!/(n-2)! = 30

n.(n-1).(n-2)!/(n-2)! = 30

n^2 - n = 30

n^2 - n - 30 = 0

a = 1; b = - 1; c = - 30

/\= b^2 - 4.1.(-30)

/\= 1 + 4.30

/\ = 121

n = [ - b +/- \/ /\] / 2a

n = [ - (-1) +/- \/121] / 2.1

n = [ 1 +/- 11] / 2

n ' = [ 1+11]/2 = 12/2 = 6

n" = [1-11]/2 = - 10/2 = - 5 (descartar)

R.: n = 6

___________

7)

A)

C 7,5 = 7!/ 5! . (7-5)!

C 7,5 = 7! / 5!. 2!

C 7,5 = 7.6.5! / 5! . 2

C7,5 = 42/2

C7,5 = 21

R.: 21

_____________

B)

C n, 4 = 4 . C n,3

n! / 4!.(n-4)! = 4. n! / 3! (n-3)!

1/ 4.3! . (n-4)! = 4 / 3! . (n -3)!

1/4.(n-4)! = 4 / (n-3)!

(n-3)! / (n-4)! = 4.4

(n-3).(n-4)!/(n-4)! = 16

n - 3 = 16

n = 16+3

n = 19

_________

8)

(n+1)! / (n-1)! = 12

(n+1).n.(n-1)! / (n-1)! = 12

n^2+ n = 12

n^2 + n - 12 = 0

a = 1; b = 1; c = - 12

/\= b^2 - 4ac

/\ = 1^2 - 4.1.(-12)

/\ = 1+48

/\= 49

n = (-b +/- \/ /\] / 2a

n = (-1 +/- \/49] / 2.1

n = (-1 +/- 7)/2

n' = (-1+7)/2 = 6/2= 3

n" = (-1-7)/2 = -8/2 = - 4 (descartar)

R.: n = 3

Answer 2

Combinação simples

$\huge\boxed{\boxed{\boxed{\boxed{\mathsf{C_{n}^p=\dfrac{n!}{p!\cdot(n-p)!}}}}}}$

Arranjo simples

$\huge\boxed{\boxed{\boxed{\boxed{\mathsf{A_{n}^p=\dfrac{n!}{(n-p)!}}}}}}$

$\dotfill$

6)

a)

$\mathsf{A_{5,2}=\dfrac{5!}{(5-2)!}}\\\mathsf{A_{5,2}=\dfrac{5!}{3!}=\dfrac{5.4.3!}{3!}}\\\huge\boxed{\boxed{\boxed{\boxed{\mathsf{A_{5,2}=20}}}}}$

b)

$\mathsf{A_{n,2}=30}\\\mathsf{\dfrac{n!}{(n-2)!}</p><p> =30}\\\mathsf{\dfrac{n\cdot(n-1)\cdot(n-2)!}{(n-2)!}=30}\\\mathsf{n(n-1)=30}\\\mathsf{n^2-n-30=0}\\\mathsf{\Delta=1+120=121}\\\mathsf{n=\dfrac{1+11}{2}=\dfrac{12}{2}=6}$

$\dotfill$

7)

a)

$\mathsf{C_{7,5}=\dfrac{7!}{5!(7-5)!}}\\\mathsf{C_{7,5}=\dfrac{7\cdot6\cdot5!}{5!\cdot2!}}\\\mathsf{C_{7,5}=\dfrac{42}{2}}\\\huge\boxed{\boxed{\boxed{\boxed{\mathsf{C_{7,5}=21}}}}}$

b)

$\mathsf{C_{n,4}=4C_{n,3}}\\\mathsf{\dfrac{n!}{4!(n-4)!}=4\cdot \dfrac{n!}{3!(n-3)!}}$

$\mathsf{\dfrac{n(n-1)(n-2)(n-3)(n-4)!}{24.(n-4)!}=\dfrac{4.n(n-1)(n-2)(n-3)!}{6(n-3)!}}\\\mathsf{\dfrac{n(n-1)(n-2)(n-3)}{24}=\dfrac{2n(n-1)(n-2)}{3}}$

$\mathsf{3n(n-1)(n-2)(n-3)=48n(n-1)(n-2)}\\\mathsf{n-3=\dfrac{48n(n-1)(n-2)}{3n(n-1)(n-2)}}\\\mathsf{n-3=16}\\\mathsf{n=16+3}\\\mathsf{n=19}$

$\dotfill$

$\mathsf{\dfrac{(n+1)!}{(n-1)!}}\\\mathsf{\dfrac{(n+1)\cdot~n(n-1)!}{(n-1)!}=12}\\\mathsf{n(n+1)=12}\\\mathsf{n^2+n-12=0}\\\mathsf{a=1~~~b=1~~~c=-12}\\\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=1^2-4.1.(-12)}\\\mathsf{\Delta=1+48}\\\mathsf{\Delta=49}\\\mathsf{n=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\mathsf{n=\dfrac{-1\pm\sqrt{49}}{2\cdot1}}\\\mathsf{n=\dfrac{-1\pm7}{2}}\\\mathsf{n=\dfrac{-1+7}{2}=\dfrac{6}{2}}\\\huge\boxed{\boxed{\boxed{\boxed{\mathsf{n=3}}}}}$