BINOMIAL COEFFICIENTS (20 pts) 1. Expand the (2𝑚 − 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)

Solution.

1.

$(2x-2y)^3=8x^3-3•4x^2•2y+3•2x•4y^2-8y^3=\newline 8x^3-24x^2y+24xy^2-8y^3.$

2.

$(a-4b)^{10}=a^{10}-10a^9•4b+45a^8•(4b)^2-120a^7•(4b)^3+210a^6•(4b)^4-252a^5•(4b)^5+210a^4•(4b)^6-120a^3•(4b)^7+45a^2•(4b)^8-10a•(4b)^9+(4b)^{10} .$

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

3.

$(3x-4y)^{15}.$

The 5th term after expanding: $Т_5=С_4^{15}(3x)^{11}(4y)^{4}=1365•177147x^{11}•256y^4=1365•177147•256x^{11}y^4.$

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.