Question #194172

BINOMIAL COEFFICIENTS (20 pts)

1.    Expand the (2π‘š βˆ’ 2𝑏)3 using binomial coefficient. (14 pts)

2.    Find for the coefficient of a5b5; (a - 4b )10 (3 pts)

3.    Find the 5th term after expanding the expression (3x – 4y)15 (3 pts)

 

PIGEONHOLE PRINCIPLE

Show that in a group of 27 English words, there must be at least two that begin with the same letter. (5 pts) 

1
Expert's answer
2021-05-17T16:53:31-0400

Solution.

1.

(2xβˆ’2y)3=8x3βˆ’3β€’4x2β€’2y+3β€’2xβ€’4y2βˆ’8y3=8x3βˆ’24x2y+24xy2βˆ’8y3.(2x-2y)^3=8x^3-3β€’4x^2β€’2y+3β€’2xβ€’4y^2-8y^3=\newline 8x^3-24x^2y+24xy^2-8y^3.

2.

(aβˆ’4b)10=a10βˆ’10a9β€’4b+45a8β€’(4b)2βˆ’120a7β€’(4b)3+210a6β€’(4b)4βˆ’252a5β€’(4b)5+210a4β€’(4b)6βˆ’120a3β€’(4b)7+45a2β€’(4b)8βˆ’10aβ€’(4b)9+(4b)10.(a-4b)^{10}=a^{10}-10a^9β€’4b+45a^8β€’(4b)^2-120a^7β€’(4b)^3+210a^6β€’(4b)^4-252a^5β€’(4b)^5+210a^4β€’(4b)^6-120a^3β€’(4b)^7+45a^2β€’(4b)^8-10aβ€’(4b)^9+(4b)^{10} .

Answer.The coefficient of a5b5:a^5b^5: -258048.

3.

(3xβˆ’4y)15.(3x-4y)^{15}.

The 5th term after expanding: Π’5=Π‘415(3x)11(4y)4=1365β€’177147x11β€’256y4=1365β€’177147β€’256x11y4.Π’_5=Π‘_4^{15}(3x)^{11}(4y)^{4}=1365β€’177147x^{11}β€’256y^4=1365β€’177147β€’256x^{11}y^4.

PIGEONHOLE PRINCIPLE

The pigeonhole principle states that if n items are put into m containers, with n>m, then at least one container must contain more than one item. We have n=27 English words and m=26 English letters. n>m,27>26, there are at least two words that begin with the same letters.


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