Kombinatorika elеmеntlari
![Kombinatorika
elеmеntlari.](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_1.png)
![Ma’ruza mashg’ulotining rejasi:
•
Ko’paytma qoidasi.
•
Takrorlanadigan
o’rinlashtirishlar
•
Takrorlanmaydigan o’rin
almashtirishlar.](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_2.png)
![](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_3.png)
![](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_4.png)
![[6]
Herbert Gintis. Mathematical Literacy for Humanists. Printed in the United States of America, 2010. 61-b.](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_5.png)
![](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_6.png)
![?????? ??????
??????
=
?????? !
( ?????? − ?????? ) !](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_7.png)
![Suppose we choose m object in succession
from a set of X distinct objects a
1 , a
2 , …, a
m ,
each time recording the choice and returning the
object to the set before making the next choice.
This gives an ordered sample of the form ( b
1 , b
2 ,
…, b
k ), where each b
i is some a
j . We call this
sampling with replacement.](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_8.png)
![Faraz qilaylik, m
elementli X ={a
1 ,a
2 ,a
3 ,…,a
m }
to’plamdan ketma-ket
elementlar tanlanmoqda,
tanlangan element to’plamga
qaytarilmaslik sharti bilan.
Bu holda k o’rinli (b
1 , b
2 ,
…,b
k ) kortej hosil bo’ladi va
bu yerda har bir b
i biror a
j ga
teng bo’ladi. 1
1
Herbert Gintis. Mathematical Literacy for Humanists. Printed in the United States of America, 2010. 61-b.](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_9.png)
![Takrorlanmaydigan o’rin almashtirishlar.
1. Agar chekli X to’plam elementlari biror usul bilan
nomerlab chiqilgan bo’lsa, X to’plam tartiblangan
deyiladi.
Masalan, X= { x
1 , x
2 ,…,x
m }. Bitta to’plamni turli usullar
bilan tartiblash mumkin.
Masalan, sinf o’quvchilarini yoshiga, bo’yiga,
ogirligiga
qarab yoki o’quvchilar familiyalari bosh harflarini alifbo
bo’yicha tartiblash mumkin.
m elementli X to’plamni necha xil usul bilan tartiblash
mumkin degan savolga javob beraylik.](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_10.png)
![•
Tartiblash — bu elementlarni nomerlash
demakdir. 1-nomerni m ta elementning
istalgan biriga berish mumkin. Shuning
uchun 1-elementni m usul bilan, 2-elementni
1-element tanlanib bo’lgandan so’ng m -1
usul bilan tanlash mumkin va hokazo, oxirgi
elementni tanlash uchun faqat bitta usul
qoladi, xolos.](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_11.png)
![Tartiblashlarning umumiy soni
m(m - 1 )(m -2)·... ·2·1 = m! ga teng.
m! — dastlabki m ta natural son
ko’paytmasi m faktorial deb
o’qiladi](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_12.png)
![•
P belgisi fransuz tilidagi “permutation”, ya’ni “o`rin almashtirish”
so`zining 1- harfidan olingan
•
Masala. 8 ta ladyani shaxmat doskasida bir-birini urmaydigan qilib necha
usul bilan joylashtirish mumkin?
•
Yechish . Ladyalar soni 8 ta.
•
O`rin almashtirishlarning ba’zi qiymatlari:
ta ’ rif bo ` yicha !](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_13.png)
![Masalan, 5!= 1·2·3·4·5 = 120,
m ! = P
m bilan belgilanadi va
takrorlanmaydigan o’rin
almashtirishlar soni deb ataladi.](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_14.png)
![Mustaqil topshiriq
•
Ko‘paytma qоidasi bilan yеchiladigan
kоmbinatоrik masalalardan namuna kеltiring.
•
1 dan 9 gacha bo‘lgan raqamlardan nеchta 5
хоnali sоn tuzish mumkin? Masala y е chimi
k о mbinat о rikaning qaysi f о rmulasi bilan
if о dalanadi?
•
ekanini isb о tlang.) ( ) ( A B n B A n ](/data/documents/619af6bc-c487-4d4a-9c12-c3e69ea053a8/page_15.png)
Kombinatorika elеmеntlari.
Ma’ruza mashg’ulotining rejasi: • Ko’paytma qoidasi. • Takrorlanadigan o’rinlashtirishlar • Takrorlanmaydigan o’rin almashtirishlar.
[6] Herbert Gintis. Mathematical Literacy for Humanists. Printed in the United States of America, 2010. 61-b.