Express $11!$ and $\binom{23}{11}$ as products of their prime factors

Question

I'm a bit stuck on how to figure this question out without a calculator and what kind of working I'm supposed to show. Any help would be appreciated, thank you. $\ddot\smile$

Factorise $11!$ and $\binom{23}{11}$ into primes without a calculator in any way. Use this to calculate their $\gcd$ and $\rm{lcm}$, and confirm how these relate to their product.

Answer 1

We have

\begin{align*} 11 &= 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\\ &= 11\times 2\times 5 \times 3^2 \times 2^3 \times 7 \times 3\times 2\times 5\times 2^2 \times 3 \times 2\\ &= 2^8 \times 3^4 \times 5^2 \times 7 \times 11, \end{align*} and \begin{align*} \binom{23}{11} &= \frac{23!}{11!12!}\\ &= \frac{23\times 22\times 21\times 20 \times 19 \times 18\times 17\times 16\times 15 \times 14 \times 13}{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\\ &= 23\times 19 \times 17\times 14 \times 13\\ &= 2\times 7\times 13\times 17 \times 19\times 23. \end{align*}

The GCD is $$2\times 7$$

and the LCM is

$$2^8 \times 3^4 \times 5^2 \times 7 \times 11\times 13\times 17 \times 19\times 23.$$

It is easily checked that if we let the two numbers be $a,b$ respectively, then $$\gcd(a,b)\times \operatorname{lcm}(a,b) = ab.$$

Answer 2

$11!=11\cdot10 \,\cdot \, ... \cdot \, 3\cdot2\cdot1$.

$10=5 \cdot2$, $9=3^2$, $8=2^3$, $6=3\cdot 2$, and $4=2^2$.

So, $11!=(11)(5 \cdot2)(3^2)(2^3)(7)(3\cdot2)(5)(2^2)(3)(2)(1)=(11)(7)(5^2)(3^4)(2^8)$.

Note that $\displaystyle\binom{23}{11}=\dfrac{23!}{11!(23-11)!}=\dfrac{23!}{11!\cdot12!}$ Can you continue to write the prime factors using the method shown above?