Almashtirish va faktorial

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16.08.2023

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Almashtirish va
faktorial
Лавлинский М.В. , LavlinskiMV@mail.ru

I. Факториал
n (n!) sonining faktoriali 1 dan n gacha bo‘lgan barcha
natural sonlarning ko‘paytmasidir.
Agar n=0 bo’lsa, n!=1
Agar n>0 bo’lsa, n!=1 2  3  …

n
#:
2! =
3! =
4! =
5! = 1  2 = 2
1  2  3 = 6
1  2  3  4 = 24
1  2  3  4  5 =
1201800yil - L. Arbogast (1759 - 1803) faktorial atamani kiritdi.
1808 yil - K. Krump (1760 - 1826) n belgisini o'ylab topdi!
Rekkurent formula:



  


0 )! 1 (
0 1
!
n n n
n
n

Faktorial jadvali
16! = 20922789888000
17! = 355687428096000
18! = 6402373705728000
19! = 121645100408832000
20! = 2432902008176640000
21! = 51090942171709440000
22! = 1124000727777607680000
23! = 25852016738884976640000
24! = 620448401733239439360000
25! = 15511210043330985984000000
26! = 403291461126605635584000000
27! = 10888869450418352160768000000
28! = 304888344611713860501504000000
29! = 8841761993739701954543616000000
30! = 2652528598121910586363084800000000! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = 40320
9! = 362880
10! = 3628800
11! = 39916800
12! = 479001600
13! = 6227020800
14! = 87178291200
15! = 1307674368000 100! ≈ 9,33×10 157
1000! ≈ 4,02×10 2567
10000! ≈ 2,85×10 35 659

IIMasala yechish
#1. Hisoblang5
! 5
5
5 ! 4 
 ! 4  4 3 2 1     24 
# 2 .
! 4 ! 3 ! 7
! 14
  4 3 2 1 3 2 1 ! 7
14 13 12 11 10 9 8 ! 7
      
      

1
14 13 12 11 5    

120120 
# 3 . 11! Faktoriali 49 ga bo’linadimi ?
7 7
11 10 9 8 7 6 5 4 3 2 1

         
Javob : Yo’q

# 4 . 26! Nechta nol bilan tugaydi 26 25 24 23 22 21
20 19 18 17 16 15 14 13 12 11
10 9 8 7 6 5 4 3 2 1
! 26
    
         
         

Javob : 6
# 5 . Kasrni qisqartiring
)! 3 4 (
)! 1 4 (


m
m
)! 3 4 (
) 1 4 )( 2 4 ( )! 3 4 (

  

m
m m m
) 1 4 )( 2 4 (    m m

#6. Ifodani soddalashtiring) 6 5 ( ! 3
) 1 3 ( )! 3 (
:
)! 3 (
! )! 3 3(
2
 
   
k k
k k
k
k k
) 1 3 ( )! 3 (
) 6 5 ( ! 3
)! 3 (
! ) 3 3( ) 2 3 ( ) 1 3 ( )! 3 (
2
 
 

      

k k
k k
k
k k k k k
) 3 )( 2 )( 1 ( !
) 6 5 ( ! 3
1
! ) 3 3 ( ) 2 3 (
2
  
 

   

k k k k
k k k k k
) 3 )( 2 )( 1 (
) 3 )( 2 ( ! 3
1
) 3 3 ( ) 2 3 (
  
 

  

k k k
k k k k
) 1 (
! 3 ) 3 3 ( ) 2 3 (

   

k
k k
) 1 (
6 ) 1 ( 3 ) 2 3 (

    

k
k k
) 2 3 ( 18    k

#7. Tenglamani yeching)! 11 ( 77 )! 10 (    k k
)! 11 (
)! 11 ( 77
)! 11 (
)! 10 (





k
k
k
k

)! 11 (
)! 11 ( 77
)! 11 (
) 10 ( )! 11 (




  
k
k
k
k k

77 ) 10 (   k
87  k

III. Almashtirish
n ta elementning almashtirilishi bir-biridan faqat
elementlarning ulardagi joylashish tartibi bilan farq qiluvchi
n ta elementning birikmasidir.
P
n belgilanishi
#1. Raqamlarning barcha mumkin bo'lgan almashtirishlarini
toping : 1, 2, 3.
1 2 3
2 1 3
3 1 2 1 3 2
2 3 1
3 2 1
O'zgartirishlar sonini topish formulasi :
P
n = n! P
3 = 3 ! = 6

IV. Masala yechish
Masala №1.
Musiqachilar qancha
turli yo'llar bilan
o'tirishlari mumkin ?
Yechish :
n=4
P
4 = 4! =24
Javob : 24“ Tergovchi maymun, eshak, echki Ha, maymoq
ayiq kvartetni o'ynashga qaror qildi lar .
Bizda notalar, bas, viola, ikkita skripka bor va
o'tloqqa o'tirdi lar ,
- O'z san'atingiz bilan nurni zabt eting dedi
maymun Ular shuncha harakat qilish sada ,
hech qanday ma'no yo'q.
"To'xtang, birodarlar, to'xtang!" - deb qichqiradi
Maymun.
Musiqa qanday ketadi? Axir siz bunday
o'tirmaysiz.
*** *** *** *** ***
Ular Eshakka bo'ysunishdi va bir qatorda
xushmuomalalik bilan o'tirishdi;
Shunday bo'lsa-da, to'rtlik yaxshi ketmaydi.
Bu erda ular har qachongidan ham ko'proq tahlil
qilishdi
Va qarama-qarshi o’tirishdi
Kimga va qanday o'tirish kerak ... "

5 ta rangli sharlarni nechi hil usul bilan grilanlarga joylashtirish
mumkin ?
Yechish :
Р
5 = 5! = 1·2·3·4·5= 120
Javob : 24Задача 2.

Masala № 3. 9-sinfning payshanba kungi jadvaliga 6 ta fan
kiritilishi kerak: rus tili, adabiyot, algebra, geografiya, fizika,
jismoniy tarbiya. Ushbu kun uchun jadvalni nechta usulda
yaratishingiz mumkin ?
P
6 = 6! = 720
Masala № 4 .
Agar siz jismoniy tarbiya darsi oxirgi bo'lishini istasangiz, bir xil 6
ta fanning jadvalini necha xil usulda tuzishingiz mumkin?
P
5 =5!= 120
Задача №5.
Rus tili va adabiyoti yonma-yon turadigan shunday jadvalni bir
xil 6 ta fandan nechta usulda tuzish mumkin?
P
5 = 5! *2 = 24 0 (1. Rus tili 2. adabiyot )

Masala 6.
4, 5, 6, 7, 8 raqamlari yordamida hammasi bir-biridan farq
qiladigan nechta turli 5 xonali sonlarni yozish mumkin?
P
5 = 5 ! = 120
Masala 7.
8 ta kitobni javonda nechta usulda joylashtirish mumkin, agar
ular orasida bitta muallifning 2 ta kitobi bo'lsa, ular har qanday
tartibga solish bilan yonma-yon turishi kerak?
P
7 = 7 ! = 5040
5040 * 2 = 10080

V. Faktorialni umumlashtirish
1. n sonning juft faktoriali
Juft n faktorial uchun :
n!! = 2 · 4 · 6 · … · n
Toq n factorial uchun :
n!! = 1 · 3 · 5 · … · n
#: Hisoblang :
10!! =
9!! =
0!! = 2 · 4 · 6 · 8 · 10 = 3840
1 · 3 · 5 · 7 · 9 = 945
1
!!n sonining juft faktorial - n!! bilan belgilanadi
-
[1, n] segmentidagi n bilan bir xil paritetga
-
ega bo'lgan barcha natural sonlarning ko'paytmasi.

2. Праймориал ( primordial )
2. n sonining boshi - n # bilan belgilanadi - n dan
oshmaydigan tub sonlarning ko'paytmasi.
11# =
12# = 2 · 3 · 5 · 7 · 11 = 2310
#: hisoblang :
7 # =
5# =
3# =
2# = 8 # = 9 # = 10 # = 2 · 3 · 5 · 7 = 210
6 # = 2 · 3 · 5 = 30
4 # = 2 · 3 = 6
= 2
#

3 .
n sonining superfaktoriali - sf (n) bilan belgilanadi -
birinchi n seperfaktorialni- 1995 yilda - N. Sloan va S.
Plowf tomonidan aniqlangan
sf(4) = 1 ! · 2! · 3! · 4! = 2 88
#: Hisoblang :
sf(3) =
sf(2) =
sf(1) =
sf(0) = 1! · 2! · 3! = 12
1! · 2! = 2
1! = 1
1
sf

Almashtirish va faktorial Лавлинский М.В. , LavlinskiMV@mail.ru

I. Факториал n (n!) sonining faktoriali 1 dan n gacha bo‘lgan barcha natural sonlarning ko‘paytmasidir. Agar n=0 bo’lsa, n!=1 Agar n>0 bo’lsa, n!=1 2  3  …  n #: 2! = 3! = 4! = 5! = 1  2 = 2 1  2  3 = 6 1  2  3  4 = 24 1  2  3  4  5 = 1201800yil - L. Arbogast (1759 - 1803) faktorial atamani kiritdi. 1808 yil - K. Krump (1760 - 1826) n belgisini o'ylab topdi! Rekkurent formula:         0 )! 1 ( 0 1 ! n n n n n

Faktorial jadvali 16! = 20922789888000 17! = 355687428096000 18! = 6402373705728000 19! = 121645100408832000 20! = 2432902008176640000 21! = 51090942171709440000 22! = 1124000727777607680000 23! = 25852016738884976640000 24! = 620448401733239439360000 25! = 15511210043330985984000000 26! = 403291461126605635584000000 27! = 10888869450418352160768000000 28! = 304888344611713860501504000000 29! = 8841761993739701954543616000000 30! = 2652528598121910586363084800000000! = 1 1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 7! = 5040 8! = 40320 9! = 362880 10! = 3628800 11! = 39916800 12! = 479001600 13! = 6227020800 14! = 87178291200 15! = 1307674368000 100! ≈ 9,33×10 157 1000! ≈ 4,02×10 2567 10000! ≈ 2,85×10 35 659

IIMasala yechish #1. Hisoblang5 ! 5 5 5 ! 4   ! 4  4 3 2 1     24  # 2 . ! 4 ! 3 ! 7 ! 14   4 3 2 1 3 2 1 ! 7 14 13 12 11 10 9 8 ! 7                1 14 13 12 11 5      120120  # 3 . 11! Faktoriali 49 ga bo’linadimi ? 7 7 11 10 9 8 7 6 5 4 3 2 1            Javob : Yo’q

# 4 . 26! Nechta nol bilan tugaydi 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 ! 26                           Javob : 6 # 5 . Kasrni qisqartiring )! 3 4 ( )! 1 4 (   m m )! 3 4 ( ) 1 4 )( 2 4 ( )! 3 4 (      m m m m ) 1 4 )( 2 4 (    m m

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