GingerAle

What you are seeing is Rom’s resignation –retroactive to the late hours of the prior week.

Solution:

V=W/Q | V = voltage in volts; W = work in Joules; Q = Charge in Coulombs  

6v = W / (4.0E-6)

Solve this for W, and include the unit of (J)oules

GA

Solution:

\(\text {Preface:}\\ \text {Normally these payments are tabulated by using a spreadsheet; however, because there are no additional }\\ \text {charges to the card, it is easy to calculate a fixed formula by adjusting the payment percentage }\\ \text {rate of (2.0%) by the periodic interest rate.}\\ \text {The monthly interest rate is $\frac{21.0\%} {12} = 1.75\% \; (0.0175)$}\\ \text {The balance before the payment is found by multiplying the previous month's balance by (1.0175). }\\ \text {The new balance after payment is 98% of the previous balance and accrued interest. }\\ \text {Balance after payment is $(B * (1.0175 * 0.98))$ or $(B * (0.99715)$. The adjusted fixed payment } \\ \text {rate complement is (0.99715). By taking integer multiples of the adjusted fixed payment rate }\\ \text {complement and multiplying by the starting balance, the balance for any month(t) is given by:}\\ B(t) = 2800 * (0.99715^t)\\ \)

\(\text {Using this formula, solve for $\lceil {t}\rceil $ when $(2800 * (0.99715^t)) = 2500$}\\ \dfrac {2800 \cdot \:0.99715^t}{2800}=\dfrac {2500} {2800}\\ 0.99715^t=\dfrac{25}{28}\\ \log \left(0.99715^t\right)=\log \left(\dfrac{25}{28}\right)\\ t \log \left(0.9972\right)=\ log \left(\frac{25}{28}\right)\\ t=\frac {\log \tiny \left(\dfrac{25}{28}\right)}{\log \left(0.99715\right)}\\ t = 39.71 \text { and } \lceil {t}\rceil = 40\\ \text { }\\ \text {After the 40th payment, the balance will be below 2500 dollars.}\\ \)

GA

Yes, I agree. The counts increase by a factor of (3!) when the persons are distinguishable.

Other points of interest:

The set counts remain the same whether or not the chairs are distinguishable by label.  Though the chairs are distinguishable, they are “fixed” in place and there is no permutation or rotation of the chairs themselves. The chairs are chosen by sets such that that no two persons are adjacent when occupying them and rotations of the sets through the 20 chairs occur in the selection process. 

However, If the chairs are not labeled, then the distinguishable set counts of 800 drop to 40 (by factoring out the 20 rotations), because the positions of the persons are no longer discernable with respect to the chairs. Only the relative space (empty chairs) between the persons is discernable.    

Happy Birthday!!  

Solution:

\(\text{The chairs are in a circle, identified by a numeric sequence, and chair one (1) follows chair twenty (20).}\\\text{ }\\ \text{There are 20 ways to choose a sequence of three (3) consecutive chairs.}\\ \text{There are 20 ways to choose a sequence of two (2) consecutive chairs.}\\ \text{There are 16 ways to choose a third chair that is $\bf{not}$ in sequence with the two (2) consecutive chairs.}\\ \text{There are nCr(20,3) ways to choose a set of three (3) chairs.}\\ \text{So ... }\\ \dbinom {20}{3} - 20 - (20*16) = 800\\ \text{There are 800 ways to place three (3) persons in a circle of twenty chairs, such that no two persons are adjacent.}\\ \)

GA

The problem with reducing the count by dropping one of the letters is it reduces the sample space. Each letter switches place with every other letter multiple times (half of all counts) as the entropy decreases to zero. Each E will switch places with every other letter and all of these events are unique, except for when the E’s switch with each other. 

Dropping the count by one (1) and removing the E, means the one of the E’s will not switch with any other letter. By keeping the E and its counts with all letters, it remains only to factor out the redundancies caused by switches with its twin.

This brings up an interesting point: There are times when it should not be factored out.

I note the title to your post is “Scrabble probability questions” (Great descriptive title. It’s one of the reasons I answered your post.) There is no probability question here, but counting is a precursor to probability. When counting for probability the redundancies are always counted.

Consider these questions: John randomized his seven (7) tiles and creates a word.

What’s the probability the word starts with Z?   Answer: (1/7)

What’s the probability the word starts with E? Answer: (2/7)

It’s easy to see that it’s counted here, and it’s always counted in combination and permutation problems related to probability. 

GA

You should use the formula. This will help to give you an “intuitive” understanding of how the TVM works. Save the online financial calculators for checking your work. If you just use financial calculators or software to solve these, it will alter your DNA, and you’ll become a Blarney Banker.

\(\text {Here is the formula ...}\\ \text { }\\ \dfrac{FV*\left(\dfrac{r}{n}\right)} {\left(1+\dfrac{r}{n}\right)^{nt} -1} =D\\ \text{ }\\ \small \dfrac{3000000*\left(\dfrac{0.085}{12}\right)} {\left(1+\dfrac{0.085}{12}\right)^{(12*40)} -1} = 742.82\\ \text{ }\\ \small \text {A monthly deposit of 742.82 dollars for 40 years and the accrued compounded interest will have a future value of 3,000,000 dollars. }\\\)

Paste this into the forum’s calculator:

f=3000000;r=0.085;n=12;t=40; f*(r/n)/ ((1+r/n)^(n*t) -1)

GA