Prime factorizations of $11!$ and $\binom{23}{11}$

Question

Express the numbers 11! = 39,916,800 and the binomial coefficient ( 23 11 ) , each as products of their prime factors. Do this without using your calculator in any way. Use this to calculate the greatest common divisor, and least common multiple of the two numbers. Confirm how these relate to the product of the numbers.

Answer 1

$11! = 11\cdot10\cdot9\cdots 2$ what are the prime factors of the numbers you are multiplying? $(11)(2\cdot5)(3^2)(2^3)(7)\cdots$

and keep simplifying.

Answer 2

I'll do the first bit, and then you can do the second.

$11!$ by definition is $11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1$. Simply express those numbers as products of primes to find $11!=11*2*5*3*3*2*2*2*7*3*2*5*2*2*3*2=2^8*3^4*5^2*7*11$.

Now $ 23 \choose 11$ is $\frac{23!}{11!12!}$. Cancelling the $12!$ gives $\frac{23*22*21*20*19*18*17*16*15*14*13}{11!}$. We repeat what we did for $11!$ to find that $23 \choose 11$ is $\frac{23*2*11*3*7*2*2*5*19*2*3*3*17*2*2*2*2*3*5*2*7*13}{11!} = \frac{2^9*3^4*5^2*7^2*11*13*17*19*23}{11!}$. Then we use our expression from earlier to cancel some powers to find that $ 23 \choose 11$ $ =2*7*13*17*19*23$.

This is enough to work out the second half of your question, my two bits of advice for the future would be a) when given factorials, think of them as products of smaller numbers, instead of the giant number a calculator tells you and b) When posting questions, try to be a bit more specific with your title; it helps to get people looking. Hopefully this helps.

Answer 3

Let $[x]$ denote the largest integer not exceeding $x.$ For prime $p,$ the largest power of $p$ that divides $n!$ is $\;p^k\;, $ where $$k=\sum_{j=1}^{\infty}[np^{-j}].$$ Note that only finitely many of the terms in the summation are non-zero.

This makes it easy to factor $n!$ and $\binom {n}{m},$ provided that you know all of the primes not less than $n.$

For example $[23/2]+[23/4]+[23/8]+[23/16]=19$ and $[11/2]+[11/4]+[11/8]=8$ and $[12/2]+[12/4]+[12/8]=10 .$ So $2=2^{19-8-10}$ divides $\binom {23}{11}=\frac {23!}{11!12!},$ but $4$ does not.

Answer 4

By definition:

$$11!=11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$$

$$11!=11\cdot2\cdot5\cdot3^2\cdot2^3\cdot2\cdot3\cdot5\cdot2^2\cdot3\cdot2$$

$$11!=11\cdot7\cdot5^2\cdot3^4\cdot2^8$$

Similarly:

$${23\choose11}=\frac{23\cdot22\cdot21\cdot20\cdot19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13}{11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$$

$${23\choose11}=\frac{23\cdot2\cdot11\cdot3\cdot7\cdot2^2\cdot5\cdot19\cdot2\cdot3^2\cdot17\cdot2^4\cdot3\cdot5\cdot2\cdot7\cdot13}{11\cdot7\cdot5^2\cdot3^4\cdot2^8}$$

$${23\choose11}=\frac{23\cdot19\cdot17\cdot13\cdot11\cdot7^2\cdot5^2\cdot3^4\cdot2^9}{11\cdot7\cdot5^2\cdot3^4\cdot2^8}$$

$${23\choose11}=23\cdot19\cdot17\cdot13\cdot7\cdot2$$