Question

A flight of stairs has 10 steps. A person can go up the steps one at a time, two at a time or any combination of 1’s and 2’s. Find the total number of ways in which the person can go up the stairs.

Answer 1

Hint: In this problem, we are given that a flight of stairs has 10 steps in which a person can go up the steps one at a time, two at a time or any combination of 1’s and 2’s, we have to find the total number of ways in which the person can go up the stairs. Here we can use factorial methods by calculating the single and double step one by one and we can add them to get the total number of ways in which the person can go up the stairs.

Complete step by step solution:
We are given that a flight of stairs has 10 steps in which a person can go up the steps one at a time, two at a time or any combination of 1’s and 2’s.
Here we have to find the total number of ways in which the person can go up the stairs.
As there are 10 steps, we have the following possibilities.
For 10 steps, we have 1.
We can now see, for 1 double step and 8 single steps, i.e. total of 9 steps, we have
$$\Rightarrow {\displaystyle \frac{9!}{8!}}={\displaystyle \frac{9\times 8!}{8!}}=9$$
For 2 double steps and 6 single steps, i.e. total of 8 steps, we have
$$\Rightarrow {\displaystyle \frac{8!}{2!6!}}={\displaystyle \frac{8\times 7\times 6!}{2\times 6!}}=28$$
Now we can assume 3 double steps and 4 single steps, i.e. total of 7 steps, we have
$$\Rightarrow {\displaystyle \frac{7!}{4!3!}}={\displaystyle \frac{7\times 6\times 5\times 4!}{4!\times 6}}=35$$
Now for 4 double steps and 2 single steps, i.e. total of 6 steps, we have
$$\Rightarrow {\displaystyle \frac{6!}{4!2!}}={\displaystyle \frac{30}{2}}=15$$
For 5 double steps, we have 1.
We can now add these possibilities, we get
$$\Rightarrow 1+9+28+35+15+1=89$$
Therefore, the total number of ways in which the person can go up the stairs is 89.

Note: We should always remember that a factorial is a function that multiplies a number by every number below it or to find the number of ways ‘n’ objects can be arranged. Here we can analyse the answer only after adding every possibility that we have got to get the final solution.