Question

There are four basketball players $A,B,C,D.$ Initially, the ball is with $A.$ The ball is always passed from one person to a different person. In how many ways can the ball come back to $A$ after seven passes?

(For example, $A\to C\to B\to D\to A\to B\to C\to A$ and $A\to D\to A\to D\to C\to A\to B\to A$ are two ways in which the ball can come back to $A$ after seven passes)

• A. 646
• B. 446
• C. 546
• D. None of the above

Answer 1

The correct option is C 546

Let $x_n$ be the number of ways in which A can get back the ball after n passes. Let $y_n$ be the number of ways in which the ball goes back to a fixed person other than A after n passes. Then
$x_n=3y_{n-1}$,
and
$y_n=x_{n-1}+2y_{n-1}$.
We also have $x_1=0,x_2=3,y_1=1and{y}_2=2$.
Eliminating yn and $yn-1$, we get $x_{n+1}=3x_{n-1}+2x_n$. Thus
$x_3=3x_1+2x_2=2\times 3=6$;
$x_4=3x_2+2x_3=(3\times 3)+(2\times 6)=9+12=21$;
$x_5=3x_3+2x_4=(3\times 6)+(2\times 21)=18+42=60$;
$x_6=3x_4+2x_5=(3\times 21)+(2\times 60)=63+120=21$;
$x_7=3x_5+2x_6=(3\times 60)+(2\times 183)=180+366=546$.
Alternate solution: Since the ball goes back to one of the other 3 persons, we have
$x_n+3y_n=3^n$;
since there are $3^n$ ways of passing the ball in n passes. Using $x_n=3y_{n-1}$, we obtain $x_{n-1}+x_n=3^{n-1}$;with $x_1=0$. Thus
$x_7=3^6-x_6=3^6-3^5+x_5=3^6-3^5+3^4-x_4=3^6-3^5+3^4-3^3+x_3$
$=3^6-3^5+3^4-3^3+3^2-x^2=3^6-3^5+3^4-3^3+3^2-3$
$=(2\times 3^5)+(2\times 3^3)+(2\times 3)=486+54+6=546$.