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These are just standard Q's where we're arranging objects that have more than one of a kind, so we have to divide by factorials of the number of each kind. So the answers are: b) 6!/(2!*2!*2!) c) 6!/(3!*3!) d) 6!/(3!*2!*1!).

For ascending order, there is a one-to-one correspondence with the descending order permutations (by flipping the digits around), except that the ones in the descending order with 0 at the end (Case 1 above) won't actually count since that'd give us a permutation with 0 at the start, which we...

For descending order: Case 1. Our quintuple of digits contains a 0. The 0 has to go at the end, and we need to pick four more numbers from nine available and place them in descending order. This is equal to the no. of ways to just select four numbers, since each selection has precisely one...

Here's how to set it up. $\noindent I'm assuming $s$ is in km/h. The time of the trip when the average speed is $s$ is $\frac{1500}{s}$ (in hours). So the total cost is $\frac{1500}{s}\times \left(s^2 + 9000\right)$ cents. So just find the $s^{\star} >0$ that minimises this function, and...

They got their first set of three equations from the fact that alpha, beta, gamma are roots. They got the next equation by adding those three equations and rearranging. This allowed the previous results to be utilised.

Re: First Year Uni Calculus Marathon

Hint: the given inequality with the absolute values says that the distance of x from -2 plus its distance from 3 is equal to 5. This clearly happens when and only when x lies between -2 and 3 (inclusive).

Re: First Year Linear Algebra Marathon $\noindent Like, if you know it for $\mathbb{C}$, would you need much real work to get to arbitrary $\mathbb{F}$? If $A = (a_{ij})$ and the $a_{ij}$ are arbitrary complex numbers, and we assume the Cayley-Hamilton Theorem for $\mathbb{C}$, isn't the...

Re: First Year Linear Algebra Marathon If you managed to prove the Cayley-Hamilton theorem for complex matrices using a property of complex matrices (like that diagonalisable complex matrices form a dense set), would this automatically also imply the theorem for any field (or commutative ring)...

Yeah if p is a prime, then that's a true result.

$\noindent This is not true. For example, $4$ divides $2^{2}$, but $4$ does not divide $2$.$

$\noindent Almost, but these miss out on the odd numbers bigger than $\frac{50}{2} = 25$. These odd numbers $27,29,31,\ldots, 49$ (there are $\color{blue}{12}$ of them) each have just themselves in their set $S_{m}$. So the sets you wrote $S_{1}, S_{3}, S_{5},\ldots, S_{25}$ (there are...

$\noindent To get started: numbers $1,2,4,\ldots 32$ are $2^{k}\cdot 1$, $k=0,1,\ldots,5$. With $m=3$, they are $3,6,12,\ldots, 48$, i.e. $2^{k}\cdot 3$, for $k = 0,1,2,3, 4$. With $m=5$, they are $5,10,20,40$, i.e. $2^{k}\cdot 5$, with $k=0,1,2,3$. The next bunch is $2^{k}\cdot 7$, with $k =...

The right-hand matrix's columns (which started off being the standard basis vectors) are all leading columns if placed next to the reduced row-echelon form of A. Hence none of the standard basis vectors are in the image of T. (It's like augmenting A with a single vector to solve a system of...

"Implies" is the "if...then" (these two are the same thing) and should be written with a "⇒" arrow if you use an arrow. "If and only if" is different (as it goes in both directions) and is denoted with an "⟺" arrow. The "⟺" arrow can't be used for "implies", and the "⇒" arrow can't be used to...

To say this with an arrow, we should use an arrow like this: ⇒. This arrow means "if ... then" (i.e. "implies"). The one you used (double-sided arrow) instead means "if and only if" (makes sense, since that arrow goes both ways).

Possibly your intentions may have been valid, but I'm not exactly sure what you meant by your wordings. Like what did you mean when you wrote for example "Either x ∈ A ⟺ x ∈ C" near the beginning? Note also that C doesn't have to be the universal set, but it turns out that C just has to be a...

Ah right yeah, didn't read it carefully before. But I think your earlier steps are dodgy or strangely worded. Here's a way to do it. Assume the hypotheses about A,B,C and let x be in A. If x is also in B, then x is in A∩B, which implies it's in C (since A∩B ⊆ C). If x is not in B, then x is in...

How did you get U being a subset of C? (It would be the other way around in general.)

How did you get C being a universal set?