A general election is to be scheduled on 5 days in May such that it is not scheduled on two consecutive days. In how many ways can the 5 days be chosen to hold the election?
My approach
There are 31 days in MAY month
$5$ days are election days and 26 days are non election days. Set aside
5 days 26 non election days will create 27 places to insert 5 election days, like: first gap $\mathrm{d}_{1} \mathrm{d}_{2} \mathrm{d}_{3}$ $\mathrm{d}_4 \mathrm{d}_5 \mathrm{d}_6 \mathrm{d}_7 \mathrm{d}_8 \mathrm{d}_9 ........
\mathrm{d}_{26}$ last gap
Total 27 gaps to insert 5 days : number of ways $=27$ c 5
Am I going correctly
any another approach??? Thanks in advance!
Well, your answer seems to be correct!
U can also think like this,
There are 31 days in May. Number of election days $=5$ Number of non election days $=31-5=26$ Let the days on which election is to be held is denoted by $E$ So 31 days will look like $X_{1} E X_{2} E X_{3} E X_{4} E X_{5} E . X_{6},$ where $X_{i}^{\prime} s$ are the number of days between the election days .They will satisfy the following constraints... $$ \begin{array}{l} X_{1}+X_{2}+X_{3}+X_{4}+X_{5}+X_{6}=26 \\ \text { where } X_{1} \geq 0, X_{2} \geq 1, X_{3} \geq 1, X_{4} \geq 1, X_{5} \geq 1, X_{6} \geq 0 \end{array} $$ Now add 2 on both sides of the equation and substitute $t_{1}=1+X_{1}$, and $t_{6}=1+X_{6}$, we will get $$ t_{1}+X_{2}+X_{3}+X_{4}+X_{5}+t_{6}=28 $$ $$ \text { where } t_{1} \geq 1, X_{2} \geq 1, X_{3} \geq 1, X_{4} \geq 1, X_{5} \geq 1, t_{6} \geq 1 $$ By using generating functions or otherwise the answer is $\left(\begin{array}{l}n-1 \\ r-1\end{array}\right)=\left(\begin{array}{c}27 \\ 5\end{array}\right)$