Sequence and permutations | AIME II, 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME II, 2015 based on Sequence and permutations.

Sequence and permutations - AIME II, 2015


Call a permutation \(a_1,a_2,....,a_n\) of the integers 1,2,...,n quasi increasing if \(a_k \leq a_{k+1} +2\) for each \(1 \leq k \leq n-1\), find the number of quasi increasing permutations of the integers 1,2,....,7.

  • is 107
  • is 486
  • is 840
  • cannot be determined from the given information

Key Concepts


Sequence

Permutations

Integers

Check the Answer


Answer: is 486.

AIME II, 2015, Question 10

Elementary Number Theory by David Burton

Try with Hints


While inserting n into a string with n-1 integers, integer n has 3 spots where it can be placed before n-1, before n-2, and at the end

Number of permutations with n elements is three times the number of permutations with n-1 elements

or, number of permutations for n elements=3 \(\times\) number of permutations of (n-1) elements

or, number of permutations for n elements=\(3^{2}\) number of permutations of (n-2) elements

......

or, number of permutations for n elements=\(3^{n-2}\) number of permutations of {n-(n-2)} elements

or, number of permutations for n elements=2 \(\times\) \(3^{n-2}\)

forming recurrence relation as the number of permutations =2 \(\times\) \(3^{n-2}\)

for n=3 all six permutations taken and go up 18, 54, 162, 486

for n=7, here \(2 \times 3^{5} =486.\)

Header text

as

Header text

sds

Subscribe to Cheenta at Youtube


Gaps in Permutation | TOMATO Objective Problem 145

Problem - Gaps in Permutation (ISI Entrance)


Find in how many ways can letters of the word PESSIMISTIC be arranged such that no two S and no two I can come together along with S and I cannot come together.

  • 0
  • 20
  • 120
  • 2400

Key Concepts


Permutations

Combinatorics

Number Theory

Check the Answer


Answer: 2400

ISI Entrance, TOMATO Objective, Problem 145

Combinatorics by Brualdi.

Try with Hints


Arranging PESSIMISTIC generally in

$\frac{11!}{3!3!}$ ways

The letters P E M T C can be arranged in 5!=120 ways

Remaining 6 slots with six letters 3 S and 3 I can be arranged in $\frac{6!}{3!3!}$=20 ways. Then number of ways =2400

Subscribe to Cheenta at Youtube