Question

VA ROOOG AJUTATI MA CU EX 2 și 3 , sunt de clasa a 10 a , E URGEEENTTTT VA ROOOG

Answer 1

Răspuns:

2) a) -3

b) 0

c) a = b = 8

Explicație pas cu pas:

2)

$C_{n}^{k}$ = $\frac{n!}{k! (n-k)!}$

$A_{n}^{k}$ = $\frac{n!}{(n-k)!}$

$P_{3}$ = n!·

a) $C_{3}^{2}$ - $A_{4}^{2}$ + $P_{3}$

$C_{3}^{2}$ = $\frac{3!}{2! (3-2)!}$ = $\frac{3!}{2! (1)!}$ = $\frac{3!}{2! }$ =  $\frac{1X2X3}{1X2}$ = 3

$A_{4}^{2}$ = $\frac{4!}{(4-2)!}$ =  $\frac{4!}{(2)!}$ = $\frac{1X2X3X4}{1X2}$ = 3 x 4 = 12 (unde X este semnul pentru înmulțire)

$P_{3}$ = 3! =  1 · 2 · 3 = 6

$C_{3}^{2}$ - $A_{4}^{2}$ + $P_{3}$ = 3 - 12 + 6 = - 3

b) $C_{4}^{0}$ - $C_{4}^{1}$ + $C_{4}^{2}$ - $C_{4}^{3}$ + $C_{4}^{4}$ =

$C_{4}^{0}$ = $\frac{4!}{0! (4-0)!}$ = $\frac{4!}{0! (4)!}$ = $\frac{4!}{(4)!}$ = 1

$C_{4}^{1}$ = $\frac{4!}{1! (4-1)!}$ = $\frac{4!}{1! (3)!}$ = $\frac{4!}{(3)!}$ =$\frac{1X2X3X4}{1X2X3}$ = 4

$C_{4}^{2}$ = $\frac{4!}{2! (4-2)!}$ = $\frac{4!}{2! (2)!}$ = $\frac{4!}{(2)!}$ = $\frac{1X2X3X4}{1X2X1X2}$ = (3 x 4)/2 = 12/2 = 6

$C_{4}^{3}$ = $\frac{4!}{3! (4-3)!}$ = $\frac{4!}{3! (1)!}$ = $\frac{4!}{(3)!}$ = $\frac{1X2X3X4}{1X2X3}$ =  4

$C_{4}^{4}$ = $\frac{4!}{4! (4-4)!}$ = $\frac{4!}{4! (0)!}$ = $\frac{4!}{(4)!}$ = 1

$C_{4}^{0}$ - $C_{4}^{1}$ + $C_{4}^{2}$ - $C_{4}^{3}$ + $C_{4}^{4}$ =  1 - 4 + 6 - 4 + 1 = 0

b) a =  $C_{4}^{1}$ + $C_{4}^{3}$

de la exercițiul anterior știm că

$C_{4}^{1}$ = $\frac{4!}{1! (4-1)!}$ = $\frac{4!}{1! (3)!}$ = $\frac{4!}{(3)!}$ =$\frac{1X2X3X4}{1X2X3}$ = 4

$C_{4}^{3}$ = $\frac{4!}{3! (4-3)!}$ = $\frac{4!}{3! (1)!}$ = $\frac{4!}{(3)!}$ = $\frac{1X2X3X4}{1X2X3}$ =  4

deci a =  $C_{4}^{1}$ + $C_{4}^{3}$ = 4 + 4 = 8

b = $C_{3}^{0}$ + $C_{3}^{1}$ + $C_{3}^{2}$ + $C_{3}^{3}$

$C_{3}^{0}$ = $\frac{3!}{0! (3-0)!}$ = $\frac{3!}{0! (3)!}$ = $\frac{3!}{(3)!}$ = 1

$C_{3}^{1}$ = $\frac{3!}{1! (3-1)!}$ = $\frac{3!}{1! (2)!}$ = $\frac{3!}{(2)!}$ =$\frac{1X2X3}{1X2}$ = 3

$C_{3}^{2}$ = $\frac{3!}{2! (3-2)!}$ = $\frac{3!}{2! (1)!}$ = $\frac{3!}{(2)!}$ = $\frac{1X2X3}{1X2}$ = 3

$C_{3}^{3}$ = $\frac{3!}{3! (3-3)!}$ = $\frac{3!}{3! (0)!}$ = $\frac{3!}{(3)!}$ = 1

b = $C_{3}^{0}$ + $C_{3}^{1}$ + $C_{3}^{2}$ + $C_{3}^{3}$  = 1 + 3 + 3 + 1 = 8

deci a = b