mathematics-class-11-unit-08-chapter-02-permutation-combination-l-2-5-mzersks9zvy

##### <Success>Properties of permutations and combinations</Success>
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2.) $\quad (n-r) \cdot {}^n C_r=n \cdot{ }^{n-1} c_r$
$\text { Proof: LHS }=\frac{(n-r) \cdot n !}{r !(n-r) !}=\frac{n !}{r !(n-r-1) !}$
$=\frac{n(n-1) !}{r !(n-1-r) !} = n \cdot{ }^{n-1} C_r=\text { RHS } $
3.) $\quad r \cdot\left(\begin{array}{l}n \\ r\end{array}\right)=(n-r+1) \cdot\left(\begin{array}{c}n \\ r-1\end{array}\right), 1 \leq r \leq n $
$\text { Proof: LHS }=r \cdot\left(\begin{array}{l}n \\ r\end{array}\right)=\frac{r \cdot n !}{r !(n-r) !} $
$ =\frac{n !}{(r-1) !(n-r) !}=\frac{(n-r+1) \cdot n !}{(r-1) !(n-r+1) !} $
$=(n-r+1) \cdot{ }^n C_{r-1}=\text { RHS } $
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##### <Success>Properties of permutations and combinations</Success>
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4.) $\quad { }^{n+1} P_r={ }^n P_r+r \cdot{ }^n P_{r-1} $
$\text { RHS }=\frac{n !}{(n-r) !}+\frac{r \cdot n !}{(n-r+1) !} $
$=n !\left[\frac{(n-r+1)}{(n-r+1) !}+\frac{r}{(n-r+1) !}\right] = \frac{(n+1) !}{(n-r+1) !}$ 
$={ }^{n+1} P_r=\text { LHS }$
5.) $\quad { }^{n+1} P_r=r !+r\left({ }^n p_{r-1}+{ }^{n-1} p_{r-1}+\cdots+{ }^r p_{r-1}\right)$
Proof: Using the result of Ex. 4
$\text { RHS }= r !+r \cdot { }^r P_{r-1}={ }^r P_r+r \cdot { }^r P_{r-1}={ }^{r+1} P_r $
$ r+1 p+r \cdot{ }^{r+1} P_{r-1}={ }^{r+2} P_r+r \cdot{ }^{r+2} P_{r-1}=\cdots \cdots { }^nP_r + r \cdot { }^nP_{r-1} = { } ^{n+1} P _r$
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##### <Success>Properties of permutations and combinations</Success>
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6.) $\quad(n-r) \cdot{ }^n P_r=n \cdot{ }^{n-1} P_r$
$\text { Proof: LHS }=\frac{(n-r) \cdot n !}{(n-r) !}=n \cdot \frac{(n-1) !}{(n-r-1) !}$
$=n \cdot{ }^{n-1} P_r$.
7.) $\quad { }^n P_r=(n-r+1) \cdot{ }^n P_{r-1}$
$\text { Proof: RHS }=\frac{(n-r+1) (n!)}{(n-r+1) !}=\frac{n !}{(n-r) !}={ }^n P_r$
8.) $\quad \ { }^n {P_r}=n \cdot{ }^{n-1}P_{r-1}$
$\text { Proof: RHS }=n \cdot{ }^{n-1} P_{r-1}=\frac{n \cdot(n-1) !}{(n-r) !}=\frac{n !}{(n-r) !}$
$ = { }^nP_r$
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### <Success>Solution</Success>
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(i) The Total no. of ways of seating of Ramesh and Giri is 11 !
(ii) If Ramesh and Giri are in adjacent seats, then we can treat them one entity. So now these are 10! arrangements. However, they can interchange their places, so $2 \times 10$ ! is the total number of ways.
(iii) So Ruby occupies middle seat (ie seat no, 6)
Ramesh can be seated in $5_{C_1}=5$ ways
Giri can be seated in $5_{C_1}=5$ ways
Remaining 8 children can be seated in 8 remaining seats in 8! ways.
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### <Success>Solution</Success>
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So by $(M P)$, the total no of seating plans is
$$
5 \times 5 \times 8 \text { ! }
$$
In the above problem, in how many ways boys and girls can sit is alternate seats.
Solution So we can see here that six boys can occupy $\operatorname{six}$ (odd numbered) seats $\rightarrow$ in 6! ways
In remaning 5 (even numbered) seats, 5 girls can be seated is 5 ! ways.
So by the (MP) principle the total no. of ways in which boys 2 girls sit in alternate seats is $6 ! \times 5$ !
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### <Success>Exercise 4</Success>
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In a lottery 8 numbers are chosen from 1 to 99 as winning numbers. if all numbers match, John wins the first prize, if 7 numbers match John gets second prize and of 6 numbers match, John gets the third prize. In how many ways John can chose numbers so that he get some prize?
Solution: The number of ways of getting the first prize = 1
The number of ways of getting the second prize.
$(\frac{8}{7}) (\frac{91}{1}) = 728$
the number of ways of getting the third prize.
$(\frac{8}{6}) (\frac{91}{2})$ = $28 \times 4095$ = 114660
So by (AP), the total no of ways winning the prize =114660 + 728 + 1 = 115389.
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### <Success>Exercise 6</Success>
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In the above problem, find the number of even numbers.
Solution: If the first digit is (4): these the last digit can be from 0, 2, 6, 8 $\rightarrow$ 4 ways. 2 reaminng digit can be chosen is $ ^8{P_2}$ ways $4\times ^8{P_2} = 4\times\frac{8!}{6!} = 224.$
If the first digit is 3 or 5 $\rightarrow$ 2 ways, there the last digit can be from 0, 2, 4, 6 8 $\rightarrow$ 5 ways reamining 2 digits can be choosen is $^8{P_2} = \frac{8!}{6!}$ ways
$2 \times 5 \times 56 = 560$
By (AP) the total no. of even integers between 3000 & 6000 so that digits are not repeated is 224 + 560 = 784.
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