## - Pascal's Triangle and Factorials

For any positive integer $$n$$:
PASCAL’S TRIANGLE
$$(a+b)^0=1$$
$$(a+b)^1=1a+1b$$
$$(a+b)^2=1a^2+2ab+1b$$
$$(a+b)^3=1a^3+3a^2 b+3ab^2+1b^3$$
$$(a+b)^4=1a^4+4a^3 b+6a^2 b^2+4ab^3+1b^4$$
The triangle formed by the coefficients of each expansion is called PASCAL’S TRIANGLE.

$$n!=n(n-1)(n-2)(n-3)…(3)(2)(1)$$
For example, $$8!=8×7×6×5×4×3×2×1$$
$$0!$$ is defined as 1.
LESSON 1
Simplify $$\displaystyle {9!\over 6!}$$
SOLUTION
$$\displaystyle {9!\over 6!}={9×8×7×6×5×4×3×2×1\over 6×5×4×3×2×1}$$
$$=9×8×7=504$$

LESSON 2
Simplify $$n!-(n-2)!$$
SOLUTION
$$n!-(n-2)!$$
$$=n(n-1)(n-2)(n-3)…(3)(2)(1)-(n-2)!$$
$$=n(n-1)(n-2)!-(n-2)!$$
• Factor out $$(n-2)!$$
$$=(n-2)![n(n-1)-1]$$
$$=(n-2)!(n^2-n-1)$$

LESSON 3
Solve the equation $$\displaystyle {(n-1)!\over (n-3)!}=72$$.
SOLUTION
$$\displaystyle {(n-1)!\over (n-3)!}=72$$
$$\displaystyle {(n-1)(n-2)(n-3)…(3)(2)(1)\over (n-3)(n-4)(n-5)…(3)(2)(1)}=72$$
• Simplify by cancelling
$$(n-1)(n-2)=72$$
$$n^2-3n+2=72$$
$$n^2-3n-70=0$$
$$(n-10)(n+7)=0$$
$$n=10$$ since $$n=-7$$ in invalid

LESSON 4
Show that
$$\displaystyle {1\over(k+2)!}-{k+1\over (k+3)!}={2\over (k+3)!}$$
SOLUTION
• Express both fractions with a common denominator by multiplying the denominator and numerator by $$(k+3)$$
$$\displaystyle {1\over (k+2)!}-{k+1\over (k+3)!}⟹{1(k+3)\over(k+3)(k+2)!}-{k+1\over (k+3)!}$$
• Combine fractions and simplify
$$\displaystyle ={k+3\over (k+3)!}-{k+1\over (k+3)!}$$
$$\displaystyle ={k+3-(k+1)\over (k+3)!}$$
$$\displaystyle ={2\over (k+3)!}$$

Factorials can be used as an alternative to Pascal’s Triangle through the use of $$^nC_{r}$$ . This is a function that can be found on your calculator.
$$\displaystyle ^nC_{r} ={n\choose r}={n!\over (n-r)!r!}$$
We know that when $$n=5$$ the coefficients for the corresponding expansion according to Pascal’s Triangle are 1, 5, 10, 10, 5, 1.
These can also be determined as follows, with the use of your calculator.
$${5\choose 0}=1$$
$${5\choose 1}=5$$
$${5\choose 2}=10$$
$${5\choose 3}=10$$
$${5\choose 4}=5$$
$${5\choose 5}=1$$

LESSON 6
Show that $$^8C_2 =^8C_6$$ .
SOLUTION
$$^8C_2 =^8C_6$$
• Rewrite using factorials and simplify
$$\displaystyle {8!\over (8-2)!2!}={8!\over (8-6)!6!}$$
$$\displaystyle {8!\over 6!2!}={8!\over 2!6!}$$
Therefore $$^8C_2 =^8C_6$$.

LESSON 7
If $$\displaystyle ^{x-1}C_2={7\over2}\:^8C_1, x>1$$,
(i) show that $$x^2-3x-54=0$$
(ii) find $$x$$.
SOLUTION
$$\displaystyle ^{x-1}C_2={7\over2}\:^8C_1$$
• Rewrite using factorials and simplify
$$\displaystyle {(x-1)!\over (x-1-2)!2!}={7\over2} (8)$$
$$\displaystyle {(x-1)!\over 2(x-3)!}=28$$
$$\displaystyle {(x-1)(x-2)\over 2}=28$$
$$(x-1)(x-2)=56$$
$$x^2-3x+2=56$$
$$x^2-3x-54=0$$
$$(x-9)(x+6)=0$$
$$x=9$$ since $$x>1$$