For the profitable truck, the dealer bought it for , where . Therefore, .

For the unprofitable truck, the dealer bought it for , where . Therefore, .

Thus, the dealer's net profit is $8,000, and the dealer's net loss is $10,000.

His loss is greater than his profit, so Quantity B is larger.

To solve this problem, imagine that the original price of the laptop is $100. After the first 20% discount, the price of the laptop becomes $80. With the additional 10% reduction, we discount from $80, not $100. 10% of 80 is 8 so we must deduct $8 from $80. This gives $72 as the final price of the laptop after both reductions. Since we started at $100, the total discount is 28%.

We have to consider two cases.  First, the group containing the president, secretary, and treasurer represents a case of permutation.  Since the order matters in such a group, we can select from our initial 20 candidates 20 * 19 * 18, or 6840 possible groupings.

Following that, the at large group constitutes a case of combinations in which the order does not matter.  Since we have chosen 3 already for the first three slots, there will be 17 remaining people.  The formula for choosing 4-person sets out of 17 candidates is represented by the combination formula of this form:

17! / ((17-4)! * 4!) = 17! / (13! * 4!) = (17 * 16 * 15 * 14) / (4 * 3 * 2) = 17 * 4 * 5 * 7 = 2380

Thus, we have 6840 and 2380 possible groupings.  These can each be combined with each other, meaning that we have 6840  * 2380, or 16,279,200 potential boards.

The key points we need to remember are that order matters and that we are sampling without replacement.  This then becomes a simple permutation problem.  We have 52 cards to be chosen 5 at a time, so the answer is 52 * 51 * 50 * 49 * 48.

Quantity A says with replacement, so we have 52 ways to choose the first card, then we replace it so we again have 52 ways to choose the 2nd card, and similarly we have 52 ways to choose the 3rd card.  Therefore we have 52 * 52 * 52 ways of choosing 3 cards with replacement.

Quantity B says without replacement, so we have 52 ways to choose the first card, but then we don't put that card back in the deck so we have 51 ways to choose the second card. Again we don't put that card back, leaving 50 ways to choose the third card. Thus there are 52 * 51 * 50 ways of choosing 3 cards without replacement.

Clearly Quantity A is greater. In general, there should always be more ways to choose something with replacement than there are without replacement just as we showed above. If you already knew this, you could have picked Quantity A without the math. Notice that either way you come to the answer here, there is NO reason to finish the computations all the way through. This will save you time on many quantitative comparison problems. For example, we know that 52 * 52 * 52 is larger than 52 * 51 * 50 without actually figuring out what the two expressions equal.

There are 6 ways to choose the first rose bush, 5 ways to choose the second, 4 ways to choose the third, and 3 ways to choose the fourth. In total there are 6 * 5 * 4 * 3 = 360 ways to arrange the rose bushes.

We need to arrange the 5 balls in 5 positions: _ _ _ _ _.  The first position can be taken by any of the 5 balls. Then there are 4 balls left to fill the second position, and so on.  Therefore the number of arrangements is 5! = 5 * 4 * 3 * 2 * 1 = 120.

For this problem, since the order of the vases matters (red blue yellow is different than blue red yellow), we're dealing with permutations.

With  selections made from  potential options, the total number of possible permutations(order matters) is:

This is a combination problem of the form “15 choose 2” because the sets of handshakes do not matter in order.  (That is, “A shakes B’s hand” is the same as “B shakes A’s hand.”)  Using the standard formula we get: 15!/((15 – 2)! * 2!) = 15!/(13! * 2!) = (15 * 14)/2 = 15 * 7 = 105.

This is a matter of permutations and combinations. You could solve this using the appropriate formulas, but it is always the case that you can make more permutations than combinations for all groups of size greater than one because the order of selection matters; therefore, without doing the math, you know that B must be the answer.