four friends want to play a game. in how many ways can the friends from teams, if both team dont have name

Mathematics

1 answer:

SVETLANKA909090 [29]2 years ago

7 0

Answer:

So I’m guessing you want either how to get a name or name ideas?

Step-by-step explanation:

1. Compare the friend’s favorite things and lump them together (lets say one likes ballet and another likes swans so the name would be the Graceful Swans)

2.Pick a name that describes what you like about your friendship (lets say they like the love that the friends give to one another so lets say Loving Friends)

NAME IDEAS

Hunting Pack

Graceful Swans

Cloudy Skies

DREAM TEAM

Windy days

IF YOU WANT TO DIVIDE THE FRIENDS THEN THERE WOULD BE TWO FRIENDS ON EACH TEAM

Hope this helped

Plz mark Brainliest

Dont forget to smash that heart at the bottom <3

Have a great day!

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A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

MakcuM [25]

Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .