Questions & Answers

Question

Answers

A. ${{3}^{10}}$

B. ${{3}^{9}}$

C. ${{3}^{8}}$

D. None of those

Answer

Verified

129.3k+ views

To do the problem, we have to assume the number of one steps taken by the person as x, the number of 2 steps taken by the person as y, and the number of 3 steps taken by the person as z. As the total number of steps is 10 we can write that

$x+2y+3z=10\to \left( 1 \right)$

We have to use different values of z to get a set of (x, y, z) and apply the combinations formula to get the number of combinations for a particular set of (x, y, z).

Number of ways of arranging x number of a’s, y number of b’s and z number of c’s….. is given by $N=\dfrac{\left( a+b+c... \right)!}{a!\times b!\times c!...}\to \left( 2 \right)$

Case-1 z=1

Then we get $x+2y=7$

We can write the different possible combinations of x and y are

$\left( x,y \right)=\left( 1,3 \right),\left( 3,2 \right),\left( 5,1 \right)$

We can write the different possible ordered pairs of (x, y, z) as

$\left( x,y,z \right)=\left( 1,3,1 \right),\left( 3,2,1 \right),\left( 5,1,1 \right)$

For example consider $\left( x,y,z \right)=\left( 1,3,1 \right)$, there are 1-1 steps, 3-2 steps and 1-3 steps. Let us consider N(x, y, z) as the number of combinations for (x, y, z). Using the analogy in equation-2, we get

$N\left( 1,3,1 \right)=\dfrac{\left( 1+3+1 \right)!}{1!\times 3!\times 1!}=\dfrac{5!}{3!}=\dfrac{5\times 4\times 3!}{3!}=20$.

Similarly, applying it to other cases, we get

$N\left( 3,2,1 \right)=\dfrac{\left( 3+2+1 \right)!}{3!\times 2!\times 1!}=\dfrac{6!}{3!\times 2!}=\dfrac{6\times 5\times 4\times 3!}{3!\times 2}=60$

$N\left( 5,1,1 \right)=\dfrac{\left( 5+1+1 \right)!}{5!\times 1!\times 1!}=\dfrac{7!}{5!}=\dfrac{7\times 6\times 5!}{5!}=42$

Totally, $N(z=1)=N\left( 1,3,1 \right)+N\left( 3,2,1 \right)+N\left( 5,1,1 \right)=20+60+42=122$

Case-2 z=2

Then we get $x+2y=4$

We can write the different possible combinations of x and y are

$\left( x,y \right)=\left( 2,1 \right),\left( 0,2 \right),\left( 4,0 \right)$

We can write the different possible ordered pairs of (x, y, z) as

$\left( x,y,z \right)=\left( 2,1,2 \right),\left( 0,2,2 \right),\left( 4,0,2 \right)$

Using the analogy in equation-2, we get

$N\left( 2,1,2 \right)=\dfrac{\left( 2+1+2 \right)!}{2!\times 1!\times 2!}=\dfrac{5!}{2!\times 2!}=\dfrac{120}{4}=30$.

$N\left( 0,2,2 \right)=\dfrac{\left( 0+2+2 \right)!}{0!\times 2!\times 2!}=\dfrac{4!}{2!\times 2!}=\dfrac{24}{2\times 2}=6$

$N\left( 4,0,2 \right)=\dfrac{\left( 4+0+2 \right)!}{4!\times 0!\times 2!}=\dfrac{6!}{4!\times 2!}=\dfrac{6\times 5\times 4!}{4!\times 2}=15$

Totally, $N(z=2)=N\left( 2,1,2 \right)+N\left( 0,2,2 \right)+N\left( 4,0,2 \right)=30+6+15=51$

Case-3 z=3

Then we get $x+2y=1$

We can write the different possible combinations of x and y are

$\left( x,y \right)=\left( 1,0 \right)$

We can write the different possible ordered pairs of (x, y, z) as

$\left( x,y,z \right)=\left( 1,0,3 \right)$

Using the analogy in equation-2, we get .

$N\left( 1,0,3 \right)=\dfrac{\left( 1+0+3 \right)!}{1!\times 0!\times 3!}=\dfrac{4!}{1\times 3!}=\dfrac{4\times 3!}{3!}=4$

Totally, $N(z=3)=N\left( 1,0,3 \right)=4$

Case-4 z=0

Then we get $x+2y=10$

We can write the different possible combinations of x and y are

$\left( x,y \right)=\left( 10,0 \right),\left( 8,1 \right),\left( 6,2 \right),\left( 4,3 \right),\left( 2,4 \right),\left( 0,5 \right)$

We can write the different possible ordered pairs of (x, y, z) as

$\left( x,y,z \right)=\left( 10,0,0 \right),\left( 8,1,0 \right),\left( 6,2,0 \right),\left( 4,3,0 \right),\left( 2,4,0 \right),\left( 0,5,0 \right)$

Using the analogy in equation-2, we get

$N\left( 10,0,0 \right)=\dfrac{\left( 10+0+0 \right)!}{10!\times 0!\times 0!}=1$.

$N\left( 8,1,0 \right)=\dfrac{\left( 8+1+0 \right)!}{8!\times 1!\times 0!}=\dfrac{9!}{8!\times 1!}=9$

$\begin{align}

& N\left( 6,2,0 \right)=\dfrac{\left( 6+2+0 \right)!}{6!\times 2!\times 0!}=\dfrac{8!}{6!\times 2!}=28 \\

& N\left( 4,3,0 \right)=\dfrac{\left( 4+3+0 \right)!}{4!\times 3!\times 0!}=\dfrac{7!}{4!\times 3!}=\dfrac{7\times 6\times 5}{6}=35 \\

& N\left( 2,4,0 \right)=\dfrac{\left( 2+4+0 \right)!}{2!\times 4!\times 0!}=\dfrac{6!}{2!\times 4!}=15 \\

\end{align}$

$N\left( 0,5,0 \right)=\dfrac{\left( 0+5+0 \right)!}{0!\times 5!\times 0!}=1$

Finally, we get

$\begin{align}

& N\left( z=0 \right)=N\left( 10,0,0 \right)+N\left( 8,1,0 \right)N+\left( 6,2,0 \right)N+\left( 4,3,0 \right)N+\left( 2,4,0 \right)N+\left( 0,5,0 \right) \\

& N\left( z=0 \right)=1+9+28+35+15+1=89 \\

\end{align}$

Total number of ways is given by

$\begin{align}

& N=N\left( z=1 \right)+N\left( z=2 \right)+N\left( z=3 \right)+N\left( z=0 \right) \\

& N=122+51+4+89=266 \\

\end{align}$