Question

Va rog, putin ajutor! Combinari si aranjamente.

Answer 1

( x +10 ) ! / ( x -4) ! · ( x +10 - x +4 ) ! = ( x +10 )!  / ( 2x -10 )! ·( x +10 -2x +10 )!
( x +10 )! / ( x -4) ! · 14! = ( x +10 ) ! / ( 2x -10 )! · ( 20 -x ) !
14! · ( x -4 ) ! = ( 2x -10 ) ! · ( 20 - x ) !
conditie  x +10 ≥ 2x -10 
din  x ≤ 20    
se verifica x =6      din 14 = 20 -x   ;      14 ! · 2! = 2! · 14! 
                                  14= 2x -10 ;           x =12 
se cere    C ²₆  = 15   si   C ₁₂² = 66

301 .                  n! · ( n + 4 ) ! / n ! < 143 · ( n +2 ) !
 ( n +2 ) ! · ( n +3 ) · ( n +4) < 143 ·( n +2) ! 
numere consecutive  ( n +3 ) ·( n +4 ) < 143 
n ∈ N = { 0 ,1,2,3 .... } 
143 = 11 ·13 
n +3 =11 ; n =8 
n ∈ { 0,1,2,3,4,5,6,7,8} 

Answer 2

     
$300) \\ C_{x+10}^{x-4}=C_{x+10}^{2x-10} \\ Avem\;2\;solutii. \\ S1: \\ 2x-10=x-4 \\ 2x=x=10-4\;=\ \textgreater \ \;x=6\;=\ \textgreater \ \;C_x^2 =C_6^2= \frac{6\; \cdot \; 5}{1\; \cdot \; 2} = \frac{30}{2}=\boxed{15} \\ S2: \\ 2x-10=(x+10)-(x-4) \\ 2x-10=x+10-x+4 \\ 2x-10=14 \\ 2x=14+10 \\ 2x=24 \\ x = \frac{24}{2}=12\;=\ \textgreater \ \; C_x^2 =C_{12}^2= \frac{12\; \cdot \; 11}{1\; \cdot \; 2} = \frac{132}{2}=\boxed{66} \\ Raspuns\;corect: \;\;\boxed{a)}$


$301) \\ \frac{A_{n+4}^4}{(n+2)!}\ \textless \ \frac{143}{Pn} \;\;\;\;\;(\text{Permutari de n = n factorial}) \\ \frac{A_{n+4}^4\; \cdot \;P_n}{(n+2)!}\ \textless \ 143 \;\;\;\ \textless \ =\ \textgreater \ \;\;\;\frac{(n+4)(n+3)(n+2)(n+1)\; \cdot \;n!}{n! (n+1)(n+2)}\ \textless \ 143 \\(n+4)(n+3)\ \textless \ 143 \\ n^{2}+4n+3n+12\ \textless \ 143 \\ n^{2}+7n+12\ \textless \ 143 \\ n^{2}+7n+12-143 \ \textless \ 0 \\ n^{2}+7n-131 \ \textless \ 0 \\ \\ n_{12}= \frac{-7 \pm \sqrt{49+4\,\cdot \,131} }{2}=\frac{-7 \pm \sqrt{49+524} }{2}=\frac{-7 \pm \sqrt{49+524} }{2} = \frac{-7 \pm 23,937 }{2}$

$n_1=\frac{-7+ 23,937 }{2} = \frac{16,937 }{2}= 8,4685 \\ \\ n_2=\frac{-7- 23,937 }{2} =\frac{- 30,937 }{2}=-15,4685 \\ \text{Ecuatia atasata inecuatiei este negativa intre radacini, deoarece }\;a\ \textgreater \ 0 \\ n \;este \;numar\;natural. \\ =\ \textgreater \ \;n \in \{0; 1; 2; 3; 4; 5; 6; 7; 8 \} \\ Raspuns\;corect:\;\;\;\boxed{b)}$