Kombinatorika va Nyuton binomi

Ushbu darsda kombinatorika bilan chambarchas bog'liq bo'lgan Nyuton binomi haqida so'z yuritamiz. $(a+b)^2$ ni yodlash oson, lekin $(a+b)^7$ ni-chi? Har safar qavsni ochib o'tirmaslik uchun Nyuton ajoyib formula qoldirgan — va uning yuragida bizga tanish guruhlash sonlari yotadi.

§1. Guruhlash sonlari va Paskal uchburchagi

Avvalgi darsdan bizga $C_n^k$ tanish. Ular yana binomial koeffitsientlar deb ham ataladi, chunki aynan shular binom darajalarini ochganda paydo bo'ladi.

Ta'rif. $C_n^k = \dfrac{n!}{k!\,(n-k)!}$ soniga binomial koeffitsient deyiladi va $\binom{n}{k}$ ko'rinishida ham yoziladi.

Bu sonlarning eng chiroyli xossasi — Paskal uchburchagi. Har bir son ustidagi ikki sonning yig'indisiga teng:

n=0:            1
n=1:          1   1
n=2:        1   2   1
n=3:      1   3   3   1
n=4:    1   4   6   4   1

Savol. Nega chetki sonlar doim 1? Chunki $\binom{n}{0}=\binom{n}{n}=1$ — hech narsa tanlamaslik yoki hammasini tanlash faqat bir usulda bo'ladi.

§2. Nyuton binomi formulasi

Qoida. Ixtiyoriy natural $n$ uchun:

Ya'ni $(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \dots + \binom{n}{n}b^n$.

Formulani "o'qish" oson: $a$ ning darajasi $n$ dan 0 gacha kamayadi, $b$ ning darajasi 0 dan $n$ gacha ortadi, har bir hadning oldida esa mos binomial koeffitsient turadi.

Misol. $(a+b)^4$ ni yoying.
Yechish. Paskal uchburchagining 4-qatoridagi koeffitsientlar: $1,4,6,4,1$:

Javob. $a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$.

Misol. $(x-2)^3$ ni yoying.
Yechish. Bu yerda $b=-2$. Koeffitsientlar $1,3,3,1$:

Javob. $x^3 - 6x^2 + 12x - 8$.

§3. Umumiy had va xossalar

Ko'pincha butun yoyilma emas, faqat bitta had kerak bo'ladi. Buning uchun umumiy had formulasidan foydalanamiz.

Qoida. $(a+b)^n$ yoyilmasining $(k+1)$-hadi:

Misol. $(2x+3)^5$ yoyilmasida $x^2$ li had koeffitsientini toping.
Yechish. $x^2$ hosil bo'lishi uchun $a=2x$ ning darajasi 2 bo'lsin, ya'ni $n-k=2$, demak $k=3$:

Javob. Koeffitsient $1080$.

Binomial koeffitsientlarning yana ikki foydali xossasi bor:

Qoida. Barcha koeffitsientlar yig'indisi $2^n$ ga teng:

Bu $a=b=1$ qo'yilganda kelib chiqadi.

Eslatma. $\binom{n}{k}=\binom{n}{n-k}$ simmetriya xossasidan foydalanib, hisobni yengillashtiring: masalan $\binom{20}{18}$ ni sanashdan ko'ra $\binom{20}{2}=190$ ni sanash osonroq.

§ Lug'at

Nyuton binomi — Newton's binomial
Binomial koeffitsient — binomial coefficient
Paskal uchburchagi — Pascal's triangle
Yoyilma — expansion
Umumiy had — general term
Daraja — power (exponent)