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Kombinatorika o‘yini

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Kombinatorika o‘yini

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Kombinatorika bo‘yicha qisqa savollar

Created by

Sevinch Nomozova
Sevinch Nomozova
Uzbekistan

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Kombinatorika o‘yini
 

Kombinatorika o‘yiniOnline version

Kombinatorika bo‘yicha qisqa savollar

by Sevinch Nomozova
1

A, B, C, D to‘rt xil belgi bo‘lib, ularni 4 joyga nechta tartib bilan joylashtirish mumkin?

2

5 ta xil namunadan 3 tasini tanlashda (no’tartib) nechta yo‘l bor?

3

A yoki B harflaridan tashkil topgan yozuvlar uzunligi 3 bo‘lsa, qanchadan-qancha narsa bo‘ladi?

4

Raqamlar 0,1,2 bo‘lib, uch joyga takrorlanmas yozish uchun nechta yo‘l bor?

5

5 xil tipdan (takrorlanmas) 4 ta qolip tanlashda nechta usul bor?

6

2 ta harfdan iborat bo‘lgan 4 jaun yozuvlar (takrorlanmas yo‘q) nechta?

7

Agar 6 odam bor bo‘lib, ular qator bo‘yicha necha yo‘l joylashtirish mumkin?

8

8 ta odamdan 2 ta o‘quvchini tanlab, nechta yo‘l bilan mumkin?

9

A,A,B,B harflardan tashkil topgan 4 ta matnni nechta o‘zgaruvchisi mavjud?

10

Raqamlar B, L, O, N dan iborat 4 harfli so‘zlar takrorlanmas tartibda nechta?

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To‘rt turli belgi bo‘lganda tartiblar soni 4! (=24) ga teng.

Tanlashda tartib aniqlanmagan, kombinatsiya: C(5,3)=10.

Har bir joyga A yoki B kelishi mumkin, 2^3 = 8.

3 ta joy, takrorlanmas: 3P3 = 6.

Takrorlanmas tanlov formuli: C(5+4-1,4)=C(8,4)=70.

Har bir joy uchun 2 harf, takrorlanmas bo‘lsa 2^4=16.

6 ta odam uchun ketma-ketliklar 6! = 720.

Tanlashda tartib yo‘q, C(8,2)=28.

Kam yadrosiz bo‘lganlar: 4!/(2!2!)=6.

Harflar o‘zaro farqli bo‘lgani uchun 4! = 24.

 
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