Categories
AIME I Algebra Arithmetic Combinatorics Math Olympiad USA Math Olympiad

Arrangement of digits | AIME I, 2012 | Question 5

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Arrangement of digits.

Arrangement of digits – AIME 2012


Let B be the set of all binary integers that can be written using exactly 5 zeros and 8 ones where leading zeros are allowed. If all possible subtractions are performed in which one element of B is subtracted from another, find the number of times the answer 1 is obtained.

  • is 107
  • is 330
  • is 840
  • cannot be determined from the given information

Key Concepts


Arrangements

Algebra

Number Theory

Check the Answer


Answer: is 330.

AIME, 2012, Question 5

Combinatorics by Brualdi

Try with Hints


When 1 subtracts from a number, the number of digits remain constant when the initial number has units and tens place in 10

Then for subtraction from B requires one number with unit and tens place 10.

10 there, remaining 1 distribute any of other 11 then answer \({11 \choose 7} = {330}\).

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Categories
AIME I Algebra Arithmetic Combinatorics Math Olympiad USA Math Olympiad

Arrangement Problem | AIME I, 2012 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Arrangement.

Arrangement – AIME 2012


Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.

  • is 107
  • is 216
  • is 840
  • cannot be determined from the given information

Key Concepts


Arrangements

Algebra

Number Theory

Check the Answer


Answer: is 216.

AIME, 2012, Question 3

Combinatorics by Brualdi

Try with Hints


Here the number of ways to order the string BBBCCCFFF, such that one B is in first three positions

one C is in 4th to 6th positions, and one F is for last three positions. There are (3)(3)(3)=27 ways for first 3. Then for next two, 2 ways.

Then \((3.2)^3={216}\)ways

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AMC 8 USA Math Olympiad

Combinatorics : AMC 8 2008 Problem 14

What is Combinatorics?


Combinatorics is a field of Mathematics where we study in how many ways we can arrange some number of given objects following some certain rules of arrangements.

Try the problem


Three $\text{A’s}$, three $\text{B’s}$, and three $\text{C’s}$ are placed in the nine spaces so that each row and column contain one of each letter. If $\text{A}$ is placed in the upper left corner, how many arrangements are possible?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

AMC 8 , 2008 Problem 14

Combinatoric

6 out of 10

Mathematical Circles

Knowledge Graph


Use some hints


Lets solve the problem together !!!

lets name all the small boxes : We will call the box on the intersection of $i^{th}$ row and $j^{th}$ column $\text{Box}(i,j)$

$1$ of the three A’s is fixed in the $\text{Box}(1,1)$

Then A can not occur in the $1^{st}$ row and $1^{st}$ column.

So to place a B in the $1^{st}$ row we have 2 choices[ $\text{Box}(1,2)$ and $\text{Box}(1,3)$ ] and then only $1$ choice left to place C in the $1^{st}$ row.

Now can you think about the placements of A, B and C is the second row ??

A can not occur in the $1^{st}$ column so there are $2$ choices to place A in $2^{nd}$ row and then only one choice left to place C in $2^{nd}$ row.

Now think about the $3^{rd}$ row.

After placing all the letters in the described manner we are automatically left with only one way to place A, B and C in the third row.

So the total number of choice in $1^{st}$ row is $2$ for each of these ways there are $2$ choices for the $2^{nd}$ row and for each of these $2\times 2$ choices there is only one choice for the $3^{rd}$ row.

Hence the total number of choices : $2 \times 2 \times 1=4$.

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