Turli qiymatli in’yektiv (inyektsiya)

1-TOPSHIRIQ
1.1. To‘plamlar ustida amallar
1.1.14
U={ a, b, c,d, e, f, g,h }-universal to’plam
X={g,h,a,b}, Y={a,b,c,d}
1.X Y g,h,a,ba,b,c,da,b,c,d,g,he,fABᴖcAB C.
2.Y={a,b,c,d}={e,f,g,h}=AB CAB AB C.
3.XY={g,h,a,b}∆{a,b,c,d}={c,d,e,f}∆{a,b,c,d}={a,b,e,f}=
=AᴗBᴗCᴗAᴖBᴗ AB C.
4.XᴖY={g,h,a,b}ᴖ{a,b,c,d}={g,h,a,b}ᴖ{e,f,g,h}={a,b}=AᴗBᴗCᴗAᴖBᴗC
5.X\Y={g,h,a,b}\{a,b,c,d={c,d,e,f}\{e,f,g,h}={c,d}=BᴖAᴗCᴗCᴖAᴗB
1) 2) 3)
4) 5)

1.2.14

A/ B A/C A/ B /C A B C=AᴖBᴗAᴖCᴗAᴖ(BᴗC)ᴖAᴖBᴖC=BᴖCᴖA
3-TOPSHIRIQ
1.7. Munosabatlarni funksiyaga tekshirish
R={(4,a),(1,b),(2,d),(3,c)}
Dl(R)={1,2,3,4}⸦A
Dr(R)={a,b,c,d}⸦B
Javob:Funksiya bo’la oladi.
x ₁≠ x₂ uchun R(x₁)=R(x₂) bajarilganligi sababli R funksiyaga turli qiymatli in’yektiv (inyektsiya) deyiladi
4-TOPSHIRIQ
2. KOMBINATORIKA ELEMENTLARI.
2.1.14. Quyida berilgan sonlar nechta turli bo‘luvchilarga ega?
M=p1^a1*p2^a2 *….*pn^an son
(1 1)(2 1)....(n 1) ta umumiy bo‘luvchiga ega;
735000;
1) 73500=735*10³=147*5*10³=7²*3¹*5⁴*2³
2) (2+1)*()1+1*(4+1)*(3+1)=3* 2*5*4=120
Javob:120 ta turli bo’luvchilarga ega
5-TOPSHIRIQ

2.2.14. Olingan darsliklarning o‘zbekchalari kamida 2 ta bo‘ladigan qilib necha xil usulda tanlab olish mumkin?

Masala: Kitob javonida tasodifiy tartibda 15 ta darslik terilgan bo‘lib,
ularning 9 tasi o‘zbek tilida, 6 tasi rus tilida. Tavakkaliga 7 ta darslik olindi.

1)2 ta o’zbek 5 ta rus tilida C₉² * C₆⁵ 4) 5 ta o’zbek 2 ta rus tilida C₉⁵*C²₆

2)3 ta o’zbek 4 ta rus tilida C₉³*C⁴ 5) 6 ta o’zbek 1 ta rus tilida C₉⁶*C¹₆
3)4 ta o’zbek 5 ta rus tilida C₉⁴*C³₆ 6) 7 ta o’zbek 0 ta rus tilida C₉⁷*C⁰₆
C₉² * C₆⁵ + C₉³*C⁴₆+ C₉⁴*C³₆+ C₉⁵*C²₆+ C₉⁶*C¹₆+ C₉⁷*C⁰₆


6-TOPSHIRIQ
2.3.14. Futbol chempionatida 16 ta jamoa qatnashadi. Jamoalarning oltin, kumush,bronza medallar va oxirgi ikkita o‘rinni egallaydigan variantlari nechta bo‘ladi?
A_n^k=n*(n-1)*(n-2)*…(n-(k-1))
A₁₆⁴=16*(16-1)*(16-2)*(16-3)=16*15*14*13=43680
Javob:43680
7-TOPSHIRIQ
2.4.14. ELLIPS;
k1=1 ,k2=2, k3=1, k4=1 , k5=1
C₁₀(1,2,1,1,1)=10!/(1!*2!*1!*1!*1!)=10!/2!=3*4*5*6*7*8*9*10=1814400
8-TOPSHIRIQ
2.6. Кombinator tenglamalar
A_x2⁷ P _x5 P₅ 10P _x
( x+2)!/(x-5)!*(x-5)!=(5!-10)*x!
(x+1)*(x+2)=(120-10)
x²+3x-108=0
x₁=-12 x₂=9
Kvadrat tenglama yechimlari x1=-12 bizning shartni bajarmaydi Ø, x2=9
yechim esa kombinator tenglamamiz yechimi bo‘ladi.