Problem with an maths exercise (from Barron's book)

Test Preparation
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<p>In the barron’s book Maths IIC 10th edition
A salad bar has 7 ingredients excluding the dressing. How many different salads are possible where two salads?
(128) Solution: you can either include or exclude each of the seven ingredient in your salad, which means there are 2 choices for each ingredient according to the multiplication rule, there are 2^7=128 ways of making these yes-no choices.</p>

<p>At the beginning I thought the solution was 7! (what an idiot!!!) as though it was asked how many ways to make the same 7-ingredient-salad.
After brainstorming a little bit. I found a way to solve this problem but It’s not still the right answer according to the book.</p>

<p>My solution: Let’s say that a salad must have at least 1 ingredient. Thus we can conclude that the total number of different salads possible is the sum of the ways for each different number of ingredient (I hope I’m clear ;)):</p>

<p>(To those who might forget nCr=n!/((n-r)!r!) where n would represent the total number of ingredient available and r the number of ingredient for each salad)</p>

<p>7nCr1+7nCr2+7nCr3+7nCr4+7nCr5+7nCr6+7nCr7<br>
7+21+35+35+21+7+1=127</p>

<p>128 salad implies that there would be a salad without any ingredient: 7nCr0=1 it’s impossible</p>

<p>What do guys think about it? If I’m wrong please explain more accurately the Barron’s solution.
Thank you in advance.
Moreover if you know where there would be other mistakes, please tell me in order not to get crazy about them :)</p>

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<p>A salad with no ingredients should be excluded, unless of course salad dressing by itself constitutes a salad.</p>

<p>Your approach works. Barron’s does as well, except they forgot to subtract the salad with “0” ingredients. So 127. </p>

<p>Barron’s approach is quicker. Try the case of a mammoth salad bar with 100 ingredients.</p>

<p>That the 2 approaches are the same can be shown by expanding (1 + 1)^n. Clearly this is 2^k. It is also the sum that you’ve derived (+1). So you’ve actually proved the theorem. See Wikipedia or most advanced math books on probability: <a href=“http://en.wikipedia.org/wiki/Binomial_theorem[/url]”>http://en.wikipedia.org/wiki/Binomial_theorem&lt;/a&gt;&lt;/p&gt;

3

<p>Thank you fogcity,
For those interested:
This is not specified in Barron’s but this exercise is actually an application of the Pascal’s triangle.
We have (n/0)+(n/1)+…+(n/n-1)+(n/n)=2^k for those who might have read this thread.
[Pascal’s</a> triangle - Wikipedia, the free encyclopedia](<a href=“http://en.wikipedia.org/wiki/Pascal’s_triangle]Pascal’s”>Pascal's triangle - Wikipedia)</p>