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tia_tia [17]
3 years ago
11

How many different combinations of shirts and ties are possible if you have 4 shirts and 5 ties?

Mathematics
2 answers:
Fittoniya [83]3 years ago
8 0
20 different combinations.
Mademuasel [1]3 years ago
5 0
Well this is simple it would be 20 because it is 4 shirts with 5 options, 5 for every 1, 4 in total (5 x 4 = 20)
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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}.

5 0
2 years ago
Write the expression without radicals, using only positive exponents
Mamont248 [21]

Answer:

\sqrt[3]{a^{2}+b^{2}}=(a^{2}+b^{2})^{\frac{1}{3}}

Step-by-step explanation:

∵∛x = (x)^1/3

∴ \sqrt[3]{a^{2}+b^{2}}=(a^{2}+b^{2})^{\frac{1}{3}}

So you can replace the radicals by fractional exponents

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3 years ago
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vovangra [49]
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5 0
3 years ago
The sum of two consecutive even integers is −234. what are the two integers
dolphi86 [110]

<span>Does this help answer you question?</span>

<span>2x+<span>(2x+2)</span>=234</span>

<span>4x=232</span>

<span>x=58</span>

So, <span>2x=2<span>(58)</span>=116</span>

<span>116+118=234</span>

4 0
3 years ago
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The graph represents y = ⌈x⌉ + 1.<br><br><br> What is f(–2.75)?<br><br> –5<br> –4<br> –2<br> –1
wlad13 [49]

f(-2.75)=\lceil -2.75 \rceil+1=-2+1=-1

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3 years ago
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