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2. Cleaning for a Reason Cleans Homes for Women in Cancer ... - AOL

   www.aol.com/news/2012-10-16-cleaning-for-a...

   Cleaning for a Reason Cleans Homes for Women in Cancer Treatment Procter & Gamble supports charity with Limited Edition 'Pink' Swiffer Starter Kit, making life easier, and cleaner, for women with ...

3. 8 Things You Should Never Clean with a Swiffer - AOL

   www.aol.com/8-things-never-clean-swiffer...

   Swiffer products may be convenient and easy to use, but cleaning experts say they’re not safe for all surfaces. From marble surfaces to wooden decks, there are a few areas where you shouldn’t ...

4. Swiffer - Wikipedia

   en.wikipedia.org/wiki/Swiffer

   Swiffer is an American brand of cleaning products that is made by Procter & Gamble.Introduced in 1999, [1] the brand uses the "razor-and-blades" business model, whereby the consumer purchases the handle assembly at a low price, but must continue to purchase replacement refills and pads over the lifespan of the product.

5. Buy one, get one free - Wikipedia

   en.wikipedia.org/wiki/Buy_one,_get_one_free

   The economist Alex Tabarrok has argued, that the success of this promotion lies in the fact that consumers value the first unit significantly more than the second one. So compared to a seemingly equivalent "Half price off" promotion, they may only buy one item at half price, because the value they attach to the second unit is lower than even the discounted price.

6. Ann Taylor's Semi-Annual Sale is here — get an extra 60% off ...

   www.aol.com/lifestyle/ann-taylors-semi-annual...

   The semi-annual sale runs through the weekend both in stores and online, so you can take advantage of those steep discounts whether you visit in-person or browse on your computer.

7. Coupon collector's problem - Wikipedia

   en.wikipedia.org/wiki/Coupon_collector's_problem

   An alternative statement is: given n coupons, how many coupons do you expect you need to draw with replacement before having drawn each coupon at least once? The mathematical analysis of the problem reveals that the expected number of trials needed grows as Θ ( n log ⁡ ( n ) ) {\displaystyle \Theta (n\log(n))} .