ISI2019-MMA-27

Question

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?

• A. $\begin{pmatrix} 26 \\ 5 \end{pmatrix}$
• B. $\begin{pmatrix} 27 \\ 5 \end{pmatrix}$
• C. $\begin{pmatrix} 30 \\ 5 \end{pmatrix}$
• D. $\begin{pmatrix} 31 \\ 5 \end{pmatrix}$

Answer 1

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 $d$ .

So 31 days will look like $X_{1}dX_{2}dX_{3}dX_{4}dX_{5}d.X_{6}$ , where $X_{i}'s$ are the number of days between the election days .They will satisfy the following constraints...

$X_{1}+X_{2}+X_{3}+X_{4}+X_{5}+X_{6}= 26$

where $X_{1} \geq0$ , $X_{2} \geq1$, $X_{3} \geq1$, $X_{4} \geq1$, $X_{5} \geq1$, $X_{6} \geq0$

Now add $2$ on both sides of the equation and substitute  $Y_{1} = 1+ X_{1}$, and $Y_{6} = 1+ X_{6}$ , we will get
$Y_{1}+X_{2}+X_{3}+X_{4}+X_{5}+Y_{6}= 28$

where $Y_{1} \geq1$ , $X_{2} \geq1$, $X_{3} \geq1$, $X_{4} \geq1$, $X_{5} \geq1$, $Y_{6} \geq1$

. By using generating functions or otherwise the answer is $\binom{n-1}{r-1} = \binom{27}{5}$
So $B$ is correct

Comments

Thank you for this explanation!

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Answer 2

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 d1   d2   d3   d4   d5   d6   d7  d8   d9  d10  d11   d12   d13   d14   d15   d16  d17  d18  d19  d20  d21  d22  d23  d24   d25   d26 last  gap

Total 27 gaps to insert 5 days : number of ways = 27 c 5
So option B is the correct option

Answer 3

5 consecutive days means 4 barrier between them A | B | C | D | E that is 4 gaps between then so select 4 gap is equal to 4 day hence remaining days= 31-4=27

Now select 5 days out of 27 days

27C5....if any 2 event is consecutive then insert those selected 4 days between them

Option B is right