Salami

How many ways can we choose 5 pcs of salami if we have 6 types of salami for 10 pieces and one type for 4 pieces?

Correct answer:

x =  461

Step-by-step explanation:

$$\begin{gathered} C_5(10)=\left({\displaystyle \genfrac{}{}{0px}{}{10}{5}}\right)={\displaystyle \frac{10!}{5!(10-5)!}}={\displaystyle \frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{5\cdot 4\cdot 3\cdot 2\cdot 1}}=252 \\ {C}_5(9)=\left({\displaystyle \genfrac{}{}{0px}{}{9}{5}}\right)={\displaystyle \frac{9!}{5!(9-5)!}}={\displaystyle \frac{9\cdot 8\cdot 7\cdot 6}{4\cdot 3\cdot 2\cdot 1}}=126 \\ {C}_5(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{5}}\right)={\displaystyle \frac{8!}{5!(8-5)!}}={\displaystyle \frac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}}=56 \\ {C}_5(7)=\left({\displaystyle \genfrac{}{}{0px}{}{7}{5}}\right)={\displaystyle \frac{7!}{5!(7-5)!}}={\displaystyle \frac{7\cdot 6}{2\cdot 1}}=21 \\ {C}_5(6)=\left({\displaystyle \genfrac{}{}{0px}{}{6}{5}}\right)={\displaystyle \frac{6!}{5!(6-5)!}}={\displaystyle \frac{6}{1}}=6 \\ k=5 \\ x=\left(\genfrac{}{}{0px}{}{10}{k}\right)+\left(\genfrac{}{}{0px}{}{9}{k}\right)+\left(\genfrac{}{}{0px}{}{8}{k}\right)+\left(\genfrac{}{}{0px}{}{7}{k}\right)+\left(\genfrac{}{}{0px}{}{6}{k}\right)=252+126+56+21+6=461 \end{gathered}$$

Try another example

Showing 5 comments:

Hello

I tried using (1/120) as a base to multiply the spaces 1/5 but then I end up with too many to multiply

2 years ago  Like

Hello

It gets to a really small number!

2 years ago  Like

What_the_what

This question is very unclear. choosing 5 pieces from 6 types for 10 pieces and 1 type for 4 pieces doesn't even make sense. How do we unpack this question?

2 years ago  Like

Math student

My answer is 462
C(7,5) = 11!/ (5![6!) = 462

1 year ago  Like

Math student

Answer should be 461, since we have to subtract the selection where we selected 0,0,0,0,0,0,5 which isn't possible because the last type has only 4 pieces.

1 year ago  1 Like Like

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