Question

A vegetable has six carrot sticks four celery sticks five tomatoes and three pieces of cauliflower. in how many different orders could we eat all the vegetables?

Answer 1

There are 6,402,373,705,728 different orders in which we could eat all the vegetables.

Explanation:

To find the total number of different orders in which we could eat all the vegetables, we can use the formula for permutations of n objects:

`\sf\qquad\dashrightarrow P(n) = n!`

where:

• n is the total number of objects

In this case, we have a total of 18 vegetables (6 carrot sticks + 4 celery sticks + 5 tomatoes + 3 pieces of cauliflower). So we can calculate the permutation of 18 objects:

`\begin{aligned}\sf:\implies P(18)& =\sf 18! \\&=\sf 18 * 17 * 16 * ... * 3 * 2 * 1 \\ &= \boxed{\bold{\:\:6,402,373,705,728\:\:}}\:\:\:\green{\checkmark}\end{aligned}`

Therefore, there are 6,402,373,705,728 different orders in which we could eat all the vegetables.

Answer 2

There are 18 total pieces of vegetables, so there are 18 possible choices for the first piece, 17 possible choices for the second piece, 16 possible choices for the third piece, and so on, until there is only 1 possible choice for the last piece. Therefore, the total number of different orders we could eat all the vegetables is:

18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

which simplifies to:

18! / 0! = 6,402,373,705,728

Therefore, there are 6,402,373,705,728 different orders in which we could eat all the vegetables.