Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
5. Permutations and Combinations
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Exercise 8 Page 574

Practice makes perfect
a To expand the given binomial, we should recall the Binomial Theorem. It states that for every positive integer n, we can expand the expression (a+b)^n by using the numbers in the n^(th) row of Pascal's Triangle.
cc (a+b)^n= & _n C_0a^nb^0+ _n C_1a^(n-1)b^1 & + & ... & + & _n C_(n-1)a^1b^(n-1)+ _n C_na^0b^n In the above formula, _n C_0, _n C_1, ..., _n C_n are the numbers in the n^(th) row of Pascal's Triangle.

Row 1.3cm Pascal's Triangle 1.2cm cccccccccccc 0 & & & & & & 1 & & & & & 1 & & & & & 1 & & 1 & & & & 2 & & & & 1 & & 2 & & 1 & & & 3 & & & 1 & & 3 & & 3 & & 1 & & 4 & & 1 & & 4 & & 6 & & 4 & & 1 & 5 & 1 & & 5 & & 10 & & 10 & & 5 & & 1 Note that each number greater than 1 found in the triangle is the sum of the two numbers diagonally above it. Now consider the given binomial. ( x+ 3 )^5 We can substitute the first term for a and the second term for b using the Binomial Theorem equation and use the coefficients from the {\color{#FF0000}{5}}^\text{th} row of Pascal's Triangle.

(a+b)^n= _nC_0a^nb^0+ _nC_1a^(n-1)b^1+... + _nC_(n-1)a^1b^(n-1)+ _nC_na^0b^n
( x+ 3)^5= 1 x^5( 3)^0+ 5 x^4( 3)^1+ 10 x^3( 3)^2+ 10 x^2( 3)^3+ 5 x^1( 3)^4+ 1 x^0( 3)^5
Finally, let's simplify the expression.
1x^5(3)^0+5x^4(3)^1+10x^3(3)^2+10x^2(3)^3+5x^1(3)^4+1x^0(3)^5
â–¼
Simplify
1x^5(1)+5x^4(3)^1+10x^3(3)^2+10x^2(3)^3+5x^1(3)^4+1(1)(3)^5
1x^5(1)+5x^4(3)+10x^3(3)^2+10x^2(3)^3+5x(3)^4+1(1)(3)^5
x^5+5x^4(3)+10x^3(3)^2+10x^2(3)^3+5x(3)^4+3^5
x^5+5x^4(3)+10x^3(9)+10x^2(27)+5x(81)+243
x^5+15x^4+90x^3+270x^2+420x+243
b To expand the given binomial, we should recall the Binomial Theorem. It states that for every positive integer n, we can expand the expression (a+b)^n by using the numbers in the n^(th) row of Pascal's Triangle.
cc (a+b)^n= & _nC_0a^nb^0+ _nC_1a^(n-1)b^1 & + & ... & + & _nC_(n-1)a^1b^(n-1)+ _nC_na^0b^n In the above formula, _nC_0, _nC_1, ..., _nC_n are the numbers in the n^(th) row of Pascal's Triangle.

Row 1.3cm Pascal's Triangle 1.2cm cccccccccccc 0 & & & & & & 1 & & & & & 1 & & & & & 1 & & 1 & & & & 2 & & & & 1 & & 2 & & 1 & & & 3 & & & 1 & & 3 & & 3 & & 1 & & 4 & & 1 & & 4 & & 6 & & 4 & & 1 & 5 & 1 & & 5 & & 10 & & 10 & & 5 & & 1 Note that each number greater than 1 found in the triangle is the sum of the two numbers diagonally above it. Now consider the given binomial. (2p-q)^4 ⇔ ( 2p+( - q) )^4 We can substitute the first term for a and the second term for b using the Binomial Theorem equation and use the coefficients from the {\color{#FF0000}{4}}^\text{th} row of Pascal's Triangle.

(a+b)^n=_nC_0a^nb^0+ _nC_1a^(n-1)b^1+... + _nC_(n-1)a^1b^(n-1)+ _nC_na^0b^n
( 2p+( q))^4= 1( 2p)^4( - q)^0+ 4( 2p)^3( - q)^1+ 6( 2p)^2( - q)^2+ 4( 2p)^1( - q)^3+ 1( 2p)^0( - q)^4
Finally, let's simplify the expression.
1(2p)^4(- q)^0+4(2p)^3(- q)^1+6(2p)^2(- q)^2+4(2p)^1(- q)^3+1(2p)^0(- q)^4
â–¼
Simplify
1(2p)^4(1)+4(2p)^3(- q)^1+6(2p)^2(- q)^2+4(2p)^1(- q)^3+1(1)(- q)^4
1(2p)^4(1)+4(2p)^3(- q)+6(2p)^2(- q)^2+4(2p)(- q)^3+1(1)(- q)^4
(2p)^4+4(2p)^3(- q)+6(2p)^2(- q)^2+4(2p)(- q)^3+(- q)^4
2^4p^4+4(2)^3(p)^3(- q)+6(2)^2(p)^2(- q)^2+4(2p)(- q)^3+(- q)^4
16p^4+4(8)(p)^3(- q)+6(4)(p)^2( q^2)+4(2p)(- q^3)+q^4
16p^4-32p^3q+24p^2q^2-8pq^3+q^4