- 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
\((n-1)(n-2)=72\)
  • Solve quadratic
\(n^2-3n+2=72\)
\(n^2-3n-70=0\)
\((n-10)(n+7)=0\)
\(n=10\) 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 \(x>1\)