Math Problem Statement

10 Cookies have to be kept in 5 jars A, B, C, D, E. A jar may contain any number of cookies (including 0). What is the number of ways to distribute so that exactly 2 jars contain exactly 2 cookies?

Solution

To solve this problem, let’s analyze the requirements:

  1. Total Cookies: We have 10 cookies in total.
  2. Jars: There are 5 jars labeled A,B,C,D,A, B, C, D, and EE.
  3. Objective: Distribute the cookies so that exactly 2 jars contain exactly 2 cookies each.

Step 1: Define the Variables

Let’s denote the number of cookies in each jar as xA,xB,xC,xD,x_A, x_B, x_C, x_D, and xEx_E, where: xA+xB+xC+xD+xE=10x_A + x_B + x_C + x_D + x_E = 10 and xi0x_i \geq 0 for all jars.

Step 2: Specify Conditions for Exactly 2 Jars with 2 Cookies

We want exactly 2 jars to contain exactly 2 cookies each, meaning:

  • Two of these xix_i's should be equal to 2.
  • The other three should add up to 104=610 - 4 = 6.

Step 3: Select 2 Jars to Have 2 Cookies

The number of ways to choose 2 jars out of 5 to contain exactly 2 cookies is given by the combination: (52)=10\binom{5}{2} = 10

Step 4: Distribute Remaining 6 Cookies Among 3 Jars

We now have three jars that must contain the remaining 6 cookies. The jars can have any non-negative number of cookies, meaning we are looking for the number of non-negative integer solutions to: x1+x2+x3=6x_1 + x_2 + x_3 = 6 This is a classic "stars and bars" problem, where the number of non-negative integer solutions is given by: (6+3131)=(82)=28\binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} = 28

Step 5: Calculate the Total Number of Distributions

Since there are 10 ways to select the jars that will contain exactly 2 cookies each, and 28 ways to distribute the remaining 6 cookies among the other three jars, we multiply these values: 10×28=28010 \times 28 = 280

Final Answer

There are 280 ways to distribute the 10 cookies in 5 jars so that exactly 2 jars contain exactly 2 cookies.


Would you like more details or have any questions?

Here are 5 related questions that build on this problem:

  1. How would the solution change if we wanted exactly 3 jars to contain exactly 2 cookies?
  2. What if each jar must contain at least 1 cookie? How would that affect the number of ways?
  3. Can this method be generalized for different numbers of cookies or jars?
  4. How would the solution be affected if we wanted at least two jars to contain at least one cookie each?
  5. How does the stars and bars method apply when additional restrictions are added?

Tip: For combinatorial problems involving distributions with conditions, breaking down the conditions and using combinations with the "stars and bars" method simplifies the solution process.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Stars and Bars Method
Combinations

Formulas

Combination formula: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
Stars and Bars formula for non-negative integer solutions: \( \binom{n + r - 1}{r - 1} \)

Theorems

Combination theorem
Stars and Bars theorem

Suitable Grade Level

Grades 9-12