Lesson Progress
0% Complete

# Understand the problem

Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$, $2$, or $3$ at a time. For example, Jo could climb $3$, then $1$, then $2$. In how many ways can Jo climb the stairs? $\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

# 2010 AMC 8 Problem 25

##### Topic
Recursion – Combinatorics
Easy
Easy
##### Suggested Book
Excursion in Mathematics

Do you really need a hint? Try it first!

Try to recall the concept of recursion . Then proceed to apply in this question .  https://en.wikipedia.org/wiki/Recursion#In_mathematics
The number of ways to climb one stair is $1$. $\\$ There are $2$ ways to climb two stairs: $1$,$1$ or $2$. $\\$ For 3 stairs, there are $4$ ways: ($1$,$1$,$1$) ($1$,$2$) ($2$,$1$) ($3$) . $\\$ Try to aply this for 4 stairs in recursive manner .
For  4 stairs : After climbing 1 stair in one step  , Jo can climb the rest 3 stairs in (1 , 1 , 1) ,(1 , 2), (2, 1) ,(3) i.e. in 4 ways . $\\$ After climbing 2 stairs in one step , Jo can climb the rest 2 stairs in (1 , 1 ) and ( 2)   i.e. in 2 ways . $\\$   After climbing 3 stairs in one step , Jo can climb the rest 1 stair in  a  single step. $\\$ So to climb 4 stairs Jo can have (1+ 2 + 4) = 7 ways . Now time arrives to apply the recursion  so proceed to apply .
Why will we apply recursion ? $\\$ Notice that to climb 5 stairs , $\\$ After climbing 1 stair in one step , we again have to find how many ways Jo can follow to climb the rest 4 stairs . $\\$ After climbing 2 stairs in one step , we again have to find how many ways Jo can follow to climb the rest 3 stairs . $\\$ After climbing 3 stairs in one step ,we again have to find how many ways Jo can follow to climb the rest 2 stairs . $\\$ So the no ways Jo can have for 5 stairs is  ( 7 + 4 + 2) = 13 . $\\$ So for 6 stairs , the no ways Jo has (13 + 7 +4 ) = 24 .

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## Fly trapped inside cubical box | AMC 10A, 2010| Problem No 20

Try this beautiful Problem on Geometry on cube from AMC 10A, 2010. Problem-20. You may use sequential hints to solve the problem.

## Measure of angle | AMC 10A, 2019| Problem No 13

Try this beautiful Problem on Geometry from AMC 10A, 2019.Problem-13. You may use sequential hints to solve the problem.

## Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Try this beautiful Problem on Geometry from PRMO -2018.You may use sequential hints to solve the problem.

## Recursion Problem | AMC 10A, 2019| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-15, You may use sequential hints to solve the problem.

## Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-24, You may use sequential hints to solve the problem.

## Set of Fractions | AMC 10A, 2015| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2015. Problem-15. You may use sequential hints to solve the problem.

## Indian Olympiad Qualifier in Mathematics – IOQM

Due to COVID 19 Pandemic, the Maths Olympiad stages in India has changed. Here is the announcement published by HBCSE: Important Announcement [Updated:14-Sept-2020]The national Olympiad programme in mathematics culminating in the International Mathematical Olympiad...

## Positive Integers and Quadrilateral | AMC 10A 2015 | Sum 24

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2015. Problem-24. You may use sequential hints to solve the problem.

## Rectangular Piece of Paper | AMC 10A, 2014| Problem No 22

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2014. Problem-23. You may use sequential hints to solve the problem.

## Probability in Marbles | AMC 10A, 2010| Problem No 23

Try this beautiful Problem on Probability from AMC 10A, 2010. Problem-23. You may use sequential hints to solve the problem.