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Flura [38]
3 years ago
12

In how many ways can I distribute $6$ identical cookies and $6$ identical candies to $4$ children, if each child must receive ex

actly $3$ items?
Mathematics
2 answers:
AysviL [449]3 years ago
7 0

Answer: 44

The only choice to be made is how many gummy bears to give each child; once we've decided that, each child must receive the unique number of candies that will bring their total to 3 items. Thus, our problem reduces to counting integer solutions to a+b+c+d=6 that satisfy 0 <= a, b, c, d <= 3.

Ignoring the <= 3 restrictions, we have 9C3 = 84 solutions (which correspond to arrangements of 6 objects and 3 dividers).

Now we correct for the overcount. We must reject 5C3 = 10 solutions satisfying a >= 4, as well as 10 solutions with b, c, d >= 4. No further correction is needed, as it is not possible for more than one of a, b, c, d to be greater than or equal to 4.

Thus there are 84 - 4 x 10 = 44 ways to distribute the gummy bears, each of which forces a unique distribution of candies. There are 44 solutions.

But you guys should USE THE MESSAGE BOARD as well if you have trouble with AoPS HW problems.

(Solution from AoPS)

Hope that helped,

-sirswagger21

Olegator [25]3 years ago
3 0
Should distribute 2 cookies 1 candy and 2 candies 1 cookie I believe
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Reagan used 1,297 sq ft of wrapping paper to wrap gifts for everyone in her class. If each gift used the same amount of wrapping
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Explain how to multiply the following whole numbers 21 x 14
Lesechka [4]

Answer:

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

________

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

Step-by-step explanation:

Given

21\:\times \:14

Line up the numbers

\begin{matrix}\space\space&2&1\\ \times \:&1&4\end{matrix}

Multiply the top number by the bottom number one digit at a time starting with the ones digit left(from right to left right)

Multiply the top number by the bolded digit of the bottom number

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

Multiply the bold numbers:    1×4=4

\frac{\begin{matrix}\space\space&2&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&4\end{matrix}}

Multiply the bold numbers:    2×4=8

\frac{\begin{matrix}\space\space&\textbf{2}&1\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}

Multiply the top number by the bolded digit of the bottom number

\frac{\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}

Multiply the bold numbers:    1×1=1

\frac{\begin{matrix}\space\space&\space\space&2&\textbf{1}\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&\space\space&1&\space\space\end{matrix}}

Multiply the bold numbers:    2×1=2

\frac{\begin{matrix}\space\space&\space\space&\textbf{2}&1\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&2&1&\space\space\end{matrix}}

Add the rows to get the answer. For simplicity, fill in trailing zeros.

\frac{\begin{matrix}\space\space&\space\space&2&1\\ \space\space&\times \:&1&4\end{matrix}}{\begin{matrix}\space\space&0&8&4\\ \space\space&2&1&0\end{matrix}}

adding portion

\begin{matrix}\space\space&0&8&4\\ +&2&1&0\end{matrix}

Add the digits of the right-most column: 4+0=4

\frac{\begin{matrix}\space\space&0&8&\textbf{4}\\ +&2&1&\textbf{0}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{4}\end{matrix}}

Add the digits of the right-most column: 8+1=9

\frac{\begin{matrix}\space\space&0&\textbf{8}&4\\ +&2&\textbf{1}&0\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{9}&4\end{matrix}}

Add the digits of the right-most column: 0+2=2

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

Therefore,

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

________

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

6 0
3 years ago
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tester [92]

Answer:

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Step-by-step explanation:

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2 years ago
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Answer:

C(7,-8)

R= 8

The general formula of a circle = (x-a)+(y-b)=r²

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