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+1 vote

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?

- $\begin{pmatrix} 26 \\ 5 \end{pmatrix}$
- $\begin{pmatrix} 27 \\ 5 \end{pmatrix}$
- $\begin{pmatrix} 30 \\ 5 \end{pmatrix}$
- $\begin{pmatrix} 31 \\ 5 \end{pmatrix}$

+6 votes

Best answer

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

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

+2 votes

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

Total 27 gaps to insert 5 days : number of ways = 27 c 5

So option B is the correct option

0 votes

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

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

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