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Here is a question from my quiz:

Superior Pizza has seven vegetable ingredients and nine meat ingredients. The number of ways to select five ingredients (no doubling on ingredients) with at least one vegetable ingredient on a pizza is:

And there are multiple choice solutions:

  • $4242$
  • $4221$
  • $4368$
  • $4347$
  • None of the above.

Similarly, how many ways can we select five ingredients with at least one meat and one vegetable?

The way I tried to approach the first one (and second, as they are similar) is to take the following:

$$ {7 \choose 1} \cdot P(15 \space\space 4) $$

I did this, because I know I need 1 vegetable, and the rest can be any of the remaining $15$ ingredients.

However, I got an absurdly large answer that I don't think is correct. I don't have the correct answer yet, but I know that I got an $8/10$ on the quiz and I believe these two are the ones I did wrong. Can anyone help me solve this properly?

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3 Answers 3

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For the first question, you want combinations, not permutations, as you only care what toppings are on the pizza. If you select one vegetable first, you will double count the pizzas with two vegetables, once where each vegetable is the first one. A better approach is to count the total number of five item pizzas, then subtract the ones that have no vegetables.

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  • $\begingroup$ So would that be something like ${16 \choose 5} - {9 \choose 5}$ ? $\endgroup$
    – AdamMc331
    Commented Nov 16, 2014 at 21:24
  • $\begingroup$ Exactly correct. $\endgroup$ Commented Nov 16, 2014 at 21:38
  • $\begingroup$ I was torn on who to accept for this one, but you gave a bit of a deeper explanation so I think it should go to you. Thanks for your help. Thankfully if I do better on the test it will replace my quiz score, but 8/10 isn't too bad! $\endgroup$
    – AdamMc331
    Commented Nov 16, 2014 at 21:40
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Total number of ways of $5$ ingredients is $\dbinom{16}5$. Number of ways of no vegetable ingredient is $\dbinom{9}5$. Hence, total number of ways is $$\dbinom{16}5-\dbinom95 = \dfrac{16\times15\times14\times13\times12-9\times8\times7\times6\times5}{120} = 4242$$

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  • $\begingroup$ So if I want combinations with at least one vegetable and at least one meat, can I take the above and subtract the options with no meats/all vegetables? $\endgroup$
    – AdamMc331
    Commented Nov 16, 2014 at 21:34
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    $\begingroup$ @McAdam331 Yes. $\endgroup$
    – Adhvaitha
    Commented Nov 16, 2014 at 21:35
  • $\begingroup$ its 4242 you have a calculation mistake $\endgroup$ Commented Nov 16, 2014 at 21:44
  • $\begingroup$ @RossMillikan Yes, thanks. $\endgroup$
    – Adhvaitha
    Commented Nov 16, 2014 at 21:45
  • $\begingroup$ This is correct answer. Thank you. $\endgroup$
    – Avv
    Commented Jul 6, 2021 at 17:18
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At least one vegetable ingredient means there can be 1,2,3,4 or 5 vegetable ingredients so the answer is $\binom{7}{1}.\binom{9}{4}+\binom{7}{2}.\binom{9}{3}+\binom{7}{3}.\binom{9}{2}+\binom{7}{4}.\binom{9}{1}+\binom{7}{5}.\binom{9}{0}$=4242 (Option One).

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  • $\begingroup$ That makes sense, but your answer is slightly off from Adhvaitha's? Am I missing something here or was there possibly a miscalculation? $\endgroup$
    – AdamMc331
    Commented Nov 16, 2014 at 21:30
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    $\begingroup$ @McAdam331 its 4242 even from Adhvaithas method $\endgroup$ Commented Nov 16, 2014 at 21:44
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    $\begingroup$ Ah, he must have made a typo, he edited. Thanks for the help! $\endgroup$
    – AdamMc331
    Commented Nov 16, 2014 at 21:45

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