perms and combs question (1 Viewer)

banigul@30 · Mar 7, 2024

hi, i need help with this question, i have already solved it up to a point but i don't know where to go with it, also i think there may be any easier way to solve it
any help is appreciated!

WeiWeiMan · Mar 7, 2024

banigul@30 said:
hi, i need help with this question, i have already solved it up to a point but i don't know where to go with it, also i think there may be any easier way to solve it
any help is appreciated!

View attachment 42672

Consider the combination formula
nCr = n!/r!(n-r!)

By inspection, n=6+4=10

banigul@30 · Mar 7, 2024

WeiWeiMan said:
Consider the combination formula
nCr = n!/r!(n-r!)

By inspection, n=6+4=10

thank you!

Luukas.2 · Mar 7, 2024

WeiWeiMan said:
Consider the combination formula
nCr = n!/r!(n-r!)

By inspection, n=6+4=10\qquad

Each row of Pascal's Triangle has terms that are distributed symmetrically. In binomial coefficient form, this symmetry property can be expressed as:

$\binom{n}{r} = \binom{n}{n-r}$

$\begin{align*} \text{Applying the symmetry property to this problem yields:} \qquad \binom{n}{4} &= \binom{n}{n - 4} \\ \text{We are given that $\binom{n}{4} = \binom{n}{6}$, and hence:} \qquad \binom{n}{6} &= \binom{n}{n - 4} \\ 6 &= n - 4 \\ n &= 10 \end{align*}$

yanujw · Mar 8, 2024

Some good answers were provided in this thread. Another approach is to brute force the answer by algebra;

$\frac{n!}{6!(n-6)!}=\frac{n!}{4!(n-4)!}$
$\frac{(n-4)!}{(n-6)!}=\frac{6!}{4!}$
$(n-4)(n-5)=30$

... and solving this quadratic gives the solutions n=10 and n=-1, the latter is disregarded.

banigul@30 · Mar 8, 2024

yanujw said:
Some good answers were provided in this thread. Another approach is to brute force the answer by algebra;

$\frac{n!}{6!(n-6)!}=\frac{n!}{4!(n-4)!}$
$\frac{(n-4)!}{(n-6)!}=\frac{6!}{4!}$
$(n-4)(n-5)=30$

... and solving this quadratic gives the solutions n=10 and n=-1, the latter is disregarded.

thank you for this, this was the answer in the textbook! i tried using alegbra too but i made it too complicated haha

Luukas.2 · Mar 9, 2024

banigul@30 said:
thank you for this, this was the answer in the textbook! i tried using alegbra too but i made it too complicated haha

Your algebra is correct, you got to

$\begin{align*} \frac{n!\left[4!(n - 5)(n - 4) - 6!\right]}{(n -4)!6!4!} &= 0 \\ 4!(n - 5)(n - 4) - 6! &= 0 \qquad \text{as $\frac{n!}{(n -4)!6!4!} \neq 0$} \\ (n - 5)(n - 4) &= \frac{6!}{4!} = \frac{6 \times 5 \times 4!}{4!} \\ (n - 5)(n - 4) & = 30 \\ n &= 10 \quad \text{or} \quad -1 \\ n &= 10 \qquad \text{as $n \in \mathbb{Z}^+$} \end{align*}$

Though, rearranging in a different way is simpler:

$\begin{align*} \binom{n}{6} &= \binom{n}{4} \\ \frac{n!}{6!(n - 6)!} &= \frac{n!}{4!(n - 4)!} \\ \frac{n!}{\color{blue} 6!\color{red} (n - 6)!} \times \frac{\color{blue} 4!\color{red} (n - 4)!}{\color{black} n!} &= 1 \\ \color{blue} \frac{4!}{6!} \color{black} \times \color{red} \frac{(n - 4)!}{(n - 6)!} \color{black} \times \frac{n!}{n!} &= 1 \\ \color{blue} \frac{4!}{6 \times 5 \times 4!} \color{black} \times \color{red} \frac{(n - 4) \times (n - 5) \times (n-6)!}{(n - 6)!} \color{black} \times 1 &= 1 \\ \frac{(n - 4)(n - 5)}{30} & = 1 \qquad \text{provided $n - 6$ is a non-negative integer} \\ n^2 - 9n + 20 &= 30 \\ n^2 - 9n - 10 &= 0 \\ (n - 10)(n + 1) &= 0 \\ n &= 10 \quad \text{or} \quad -1 \\ \text{Hence, as the integer $n \ge 6$,} \qquad n &= 10 \end{align*}$

Note the application of a general and useful technique (highlighted in colour): grouping related factorials into fractions to facilitate cancelling, then unravelling whichever is the larger until the smaller is reached.

banigul@30 · Mar 9, 2024

i appreciate the help and the note on the second method, thank you!
could you explain

thanks!

Luukas.2 · Mar 9, 2024

banigul@30 said:
i appreciate the help and the note on the second method, thank you!
could you explain View attachment 42676 thanks!

Sure...

$\begin{align*} \text{Going from Line 1:} \qquad \frac{n!\left[4!(n - 5)(n - 4) - 6!\right]}{(n -4)!6!4!} &= 0 \\ \text{to Line 2:} \qquad \qquad 4!(n - 5)(n - 4) - 6! &= 0 \end{align*} \\ \text{was done by either dividing both sides of the equation by $\frac{n!}{(n -4)!6!4!}$, or multiplying both sides by} \\ \\ \text{$\frac{(n -4)!6!4!}{n!}$. In either case, to be a valid operation on an equation, the expression involved must be} \\ \\ \text{non-zero. However, since all terms within these expressions are factorials of non-negative integers,} \\ \\ \text{each expression is a ratio of positive integers, and so the Line 1 to Line 2 simplification is valid.}$

$\begin{align*} \text{Presenting from Line 1 in a different way:} \qquad \frac{n!\left[4!(n - 5)(n - 4) - 6!\right]}{(n -4)!6!4!} &= 0 \\ \frac{n!}{(n -4)!6!4!} \times \left[4!(n - 5)(n - 4) - 6!\right] &= 0 \\ \text{On multiplying by $\frac{(n -4)!6!4!}{n!} > 0$:} \qquad \frac{n!}{(n -4)!6!4!} \times \frac{(n -4)!6!4!}{n!} \times \left[4!(n - 5)(n - 4) - 6!\right] &= 0 \times \frac{(n -4)!6!4!}{n!} \\ 1 \times \left[4!(n - 5)(n - 4) - 6!\right] &= 0 \qquad \text{which is Line 2} \end{align*}$

perms and combs question (1 Viewer)

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