Question 1: Evaluate each of the following:

$\displaystyle \text{i) } ^8P_3 \hspace{1.0cm} \text{ii) } ^{10}P_4 \hspace{1.0cm} \text{iii) } ^6P_6 \hspace{1.0cm} \text{iv) } P(6, 4)$

$\displaystyle \text{We know, } ^nP_r = \frac{n!}{(n-r)!}$

$\displaystyle \text{i) } ^8P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!} = \frac{8 \times 7 \times 6 \times 5!}{5!} = 8 \times 7 \times 6 = 336$

$\displaystyle \text{ii) } ^{10}P_4 = \frac{10!}{(10-4)!} = \frac{10!}{6!} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6!} = 10 \times 9 \times 8 \times 7 = 5040$

$\displaystyle \text{iii) } ^6P_6 = \frac{6!}{(6-6)!} = \frac{6!}{0!} = 6! = 720$

$\displaystyle \text{iv) } P(6, 4) = ^6P_4 = \frac{6!}{(6-4)!} = \frac{6!}{2!} = \frac{6 \times 5 \times 4 \times 3 \times 2!}{2!} = 6 \times 5 \times 4 \times 3 = 360$

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Question 2: If $\displaystyle P (5, r) = P (6, r - 1) , \text{ find } r$

$\displaystyle P (5, r) = P (6, r - 1)$

$\displaystyle \Rightarrow \frac{5!}{(5-r)!} = \frac{6!}{(6-r+1)!}$

$\displaystyle \Rightarrow \frac{5!}{(5-r)!} = \frac{6 \times 5!}{(7-r)!}$

$\displaystyle \Rightarrow \frac{5!}{(5-r)!} = \frac{6 \times 5!}{(7-r)(6-r)(5-r)!}$

$\displaystyle \Rightarrow \frac{1}{1} = \frac{6 }{(7-r)(6-r)}$

$\displaystyle \Rightarrow (7-r)(6-r) = 6$

$\displaystyle \Rightarrow 42 - 6r - 7 r + r^2 = 6$

$\displaystyle \Rightarrow r^2 - 13 r + 36 = 0$

$\displaystyle \Rightarrow ( r-9)(r-4) = 0$

$\displaystyle \Rightarrow r = 9 \text{ or } r = 4$

Since $\displaystyle \Rightarrow r \leq n, \therefore r \neq 9$

$\displaystyle \text{Hence } \Rightarrow r = 4$

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Question 3: If $\displaystyle 5 P(4, n) = 6 \cdot P (5, n - 1) , \text{ find } n$

$\displaystyle 5 P(4, n) = 6 \cdot P (5, n - 1)$

$\displaystyle \Rightarrow 5 \cdot ^4P_n = 6 \cdot ^5P_{n-1}$

$\displaystyle \Rightarrow 5 \cdot \frac{4!}{(4-n)!} = 6 \cdot \frac{5!}{(5-n+1)!}$

$\displaystyle \Rightarrow \frac{1}{(4-n)!} = \frac{6}{(6-n)!}$

$\displaystyle \Rightarrow \frac{1}{(4-n)!} = \frac{6}{(6-n)(5-n)(4-n)!}$

$\displaystyle \Rightarrow (6-n)(5-n) = 6$

$\displaystyle \Rightarrow (6-n)(5-n) = 3 \times 2$

Comparing LHS and RHS we get

$\displaystyle 6 - n = 3 \Rightarrow n = 3$

$\displaystyle 5-n = 2 \Rightarrow n = 3$

$\displaystyle \text{Hence } n = 3$

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Question 4: If $\displaystyle P (n,5) =20 \cdot P(n, 3) , \text{ find } n$

$\displaystyle P (n,5) =20 \cdot P(n, 3)$

$\displaystyle \Rightarrow ^nP_5 = 20 \cdot ^nP_3$

$\displaystyle \Rightarrow \frac{n!}{(n-5)!} = 20 \cdot \frac{n!}{(n-3)!}$

$\displaystyle \Rightarrow \frac{1}{(n-5)!} = \frac{20}{(n-3)(n-4)(n-5)!}$

$\displaystyle \Rightarrow (n-3)(n-4) = 20$

$\displaystyle \Rightarrow (n-3)(n-4) = 5 \times 4$

Comparing LHS and RHS we get

$\displaystyle n-3 = 5 \Rightarrow n = 8$

$\displaystyle n-4 = 4 \Rightarrow n = 8$

$\displaystyle \text{Hence } n = 8$

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Question 5: If $\displaystyle ^nP_4 = 360$, find the value of $\displaystyle n$.

$\displaystyle ^nP_4 = 360$

$\displaystyle \Rightarrow \frac{n!}{(n-4)!} = 360$

$\displaystyle \Rightarrow n(n-1)(n-2)(n-3) = 360$

$\displaystyle \Rightarrow n(n-1)(n-2)(n-3) = 6 \times 5 \times 4 \times 3$

Comparing LHS and RHS we get

$\displaystyle n = 6$

$\displaystyle n-1 = 5 \Rightarrow n = 6$

$\displaystyle n-2 = 4 \Rightarrow n = 6$

$\displaystyle n-3 = 3 \Rightarrow n = 6$

$\displaystyle \text{Hence } n = 6$

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Question 6: If $\displaystyle P (9, r) = 360$, find $\displaystyle r$

$\displaystyle P (9, r) = 360$

$\displaystyle \Rightarrow ^9P_r = 360$

$\displaystyle \Rightarrow \frac{9!}{(9-r)!} = 360$

$\displaystyle \Rightarrow \frac{9!}{(9-r)!} = 9 \times 8 \times 7 \times 6$

$\displaystyle \Rightarrow \frac{9!}{(9-r)!} = \frac{9!}{5!}$

$\displaystyle \Rightarrow 9-r = 5$

$\displaystyle \Rightarrow r = 4$

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Question 7: If $\displaystyle P(11, r) = P (12, r - 1) , \text{ find } r$

$\displaystyle P(11, r) = P (12, r - 1)$

$\displaystyle \Rightarrow \frac{11!}{(11-r)!} = \frac{12!}{(12-r+1)!}$

$\displaystyle \Rightarrow \frac{1}{(11-r)!} = \frac{12}{(13-r)!}$

$\displaystyle \Rightarrow \frac{1}{(11-r)!} = \frac{12}{(13-r)(12-r)(11-r)!}$

$\displaystyle \Rightarrow (13-r)(12-r) = 12$

$\displaystyle \Rightarrow (13-r)(12-r) = 4 \times 3$

Comparing LHS and RHS we get

$\displaystyle 13-r = 4 \Rightarrow r = 9$

$\displaystyle 12 - r = 3 \Rightarrow r = 9$

$\displaystyle \therefore r = 9$

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Question 8: If $\displaystyle P (n, 4) =12 \cdot P (n,2) , \text{ find } n$

$\displaystyle P (n, 4) =12 \cdot P (n,2)$

$\displaystyle \Rightarrow \frac{n!}{(n-4)!} = 12 \cdot \frac{n!}{(n-2)!}$

$\displaystyle \Rightarrow \frac{1}{(n-4)!} = \frac{12}{(n-2)(n-3)(n-4))!}$

$\displaystyle \Rightarrow (n-2)(n-3) = 12$

$\displaystyle \Rightarrow (n-2)(n-3) = 4 \times 3$

Comparing LHS and RHS we get

$\displaystyle n-2 = 4 \Rightarrow n = 6$

$\displaystyle n-3 = 3 \Rightarrow n=6$

$\displaystyle \therefore n = 6$

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Question 9: If $\displaystyle P (n -1, 3) : P (n, 4) =1 :9 , \text{ find } n$

$\displaystyle P (n -1, 3) : P (n, 4) =1 :9$

$\displaystyle \Rightarrow \frac{(n-1)!}{(n-1-3)!} : \frac{(n)!}{(n-4)!} = 1 : 9$

$\displaystyle \Rightarrow \frac{(n-1)!}{(n-1-3)!} \times \frac{(n-4)!}{(n)!} = \frac{1}{9}$

$\displaystyle \Rightarrow \frac{1}{n-1} = \frac{1}{9}$

$\displaystyle \Rightarrow 9 = n - 1$

$\displaystyle \Rightarrow n = 10$

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Question 10: If $\displaystyle P (2n -1, n): P (2n +1, n -1) = 22:7 , \text{ find } n$

$\displaystyle P (2n -1, n): P (2n +1, n -1) = 22:7$

$\displaystyle \Rightarrow \frac{(2n-1)!}{(2n-1-n)!} : \frac{(2n+1)!}{(2n+1-n+1)! } = 22: 7$

$\displaystyle \Rightarrow \frac{(2n-1)!}{(n-1)!} : \frac{(2n+1)!}{(n+2)! } = 22: 7$

$\displaystyle \Rightarrow \frac{(2n-1)!}{(n-1)!} \times \frac{(n+2)! }{(2n+1)!} = \frac{22}{7}$

$\displaystyle \Rightarrow \frac{(2n-1)!}{(n-1)!} \times \frac{(n+2)(n+1)(n)(n-1)! }{(2n+1)(2n)(2n-1)!} = \frac{22}{7}$

$\displaystyle \Rightarrow \frac{(n+2)(n+1)(n)}{(2n+1)(2n)} = \frac{22}{7}$

$\displaystyle \Rightarrow 7 ( n+2)(n+1) = 44 ( 2n+1)$

$\displaystyle \Rightarrow 7n^2 + 14n + 7n + 14 = 88n +44$

$\displaystyle \Rightarrow 7n^2 + 21 n + 14 = 88 n + 44$

$\displaystyle \Rightarrow 7n^2 - 67n - 30 =0$

$\displaystyle \Rightarrow (n-10)(7n+3) = 0$

$\displaystyle \Rightarrow n = 10 \text{ or } n = \frac{-3}{7}$ (this is not possible)

$\displaystyle \Rightarrow \therefore n = 10$

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Question 11: If $\displaystyle P (n, 5):P (n,3)= 2:1$ , find $\displaystyle n$.

$\displaystyle P (n, 5):P (n,3) = 2:1$

$\displaystyle \Rightarrow \frac{n!}{(n-5)!} : \frac{n!}{(n-3)!} = 2:1$

$\displaystyle \Rightarrow \frac{n!}{(n-5)!} \times \frac{(n-3)!}{n!} = \frac{2}{1}$

$\displaystyle \Rightarrow \frac{(n-3)(n-4)(n-5)!}{(n-5)!} = 2$

$\displaystyle \Rightarrow (n-3)(n-4) = 2$

$\displaystyle \Rightarrow n^2 - 7n + 12 = 2$

$\displaystyle \Rightarrow n^2 - 7n + 10 = 0$

$\displaystyle \Rightarrow (n-5)(n-2) = 0$

$\displaystyle \Rightarrow n = 5 \text{ or } n = 2$ ( not possible as $\displaystyle n$ cannot be letter than 3)

$\displaystyle \therefore n = 5$

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Question 12: Prove that: $\displaystyle 1 \cdot P (1, 1) + 2 \cdot P (2, 2) + 3 \cdot P ( 3, 3) +\ldots + n \cdot P (n, n) = P (n + 1, n + 1) - 1$

LHS $\displaystyle = 1 \cdot P (1, 1) + 2 \cdot P (2, 2) + 3 \cdot P ( 3, 3) +\ldots + n \cdot P (n, n)$

$\displaystyle = 1 \cdot ^1P_1 + 2 \cdot ^2P_2 + 3 \cdot ^3P_3 +\ldots + n \cdot ^nP_n$

$\displaystyle = 1 \cdot \frac{1!}{(1-1)!} + 2 \cdot \frac{2!}{(2-2)!} + 3 \cdot \frac{3!}{(3-3)!} +\ldots + n \cdot \frac{n!}{(n-n)!}$

$\displaystyle = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! +\ldots + n \cdot n!$

$\displaystyle = \sum \limits_{r=1}^{n} r \cdot r!$

$\displaystyle = \sum \limits_{r=1}^{n} [(r+1)-1] \cdot r!$

$\displaystyle = \sum \limits_{r=1}^{n} [(r+1)! - r! ]$

$\displaystyle = [ ( 2! - 1!) + ( 3! - 2!) + \ldots + [(n+1)! - n!]$

$\displaystyle = ( n + 1 )! - 1!$

$\displaystyle = ( n + 1 )! - 1 =$ RHS. Hence proved.

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Question 13: If $\displaystyle P (15, r -1) : P (16, r - 2) = 3 : 4$, find $\displaystyle r$

$\displaystyle P (15, r -1) : P (16, r - 2) = 3 : 4$

$\displaystyle \Rightarrow ^{15}P_{r-1} : ^{16}P_{r-2} = 3:4$

$\displaystyle \Rightarrow \frac{15!}{(15-r+1)!} : \frac{16!}{(16-r+2)!} = 3:4$

$\displaystyle \Rightarrow \frac{15!}{(15-r+1)!} \times \frac{(18-r)!}{16!} = \frac{3}{4}$

$\displaystyle \Rightarrow \frac{1}{(16-r)!} \times \frac{(18-r)(17-r)(16-r)!}{16!} = \frac{3}{4}$

$\displaystyle \Rightarrow (18-r)(17-r) = 12$

$\displaystyle \Rightarrow (18-r)(17-r) = 4 \times 3$

Comparing LHS with RHS, we get

$\displaystyle 18-r = 4 \Rightarrow r = 14$

$\displaystyle 17-r = 3 \Rightarrow r = 14$

$\displaystyle \therefore r = 14$

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Question 14: If $\displaystyle ^{n+5}P_{n+1} = \Big[ \frac{11(n-1)}{2} \Big] ^{n+3}P_n$, find $\displaystyle n$

$\displaystyle ^{n+5}P_{n+1} = \Big[ \frac{11(n-1)}{2} \Big] ^{n+3}P_n$

$\displaystyle \Rightarrow \frac{(n+5)!}{(n+5-n-1)!} = \Big[ \frac{11(n-1)}{2} \Big] \frac{(n+3)!}{(n+3-n)!}$

$\displaystyle \Rightarrow \frac{(n+5)(n+4)(n+3)!}{4!} = \Big[ \frac{11(n-1)}{2} \Big] \frac{(n+3)!}{3!}$

$\displaystyle \Rightarrow \frac{(n+5)(n+4)}{4} = \frac{11(n-1)}{2}$

$\displaystyle \Rightarrow (n+5)(n+4) = 22 ( n-1)$

$\displaystyle \Rightarrow n^2 - 13n + 42 = 0$

$\displaystyle \Rightarrow (n-7)(n-6) = 0$

$\displaystyle \Rightarrow n = 7 \text{ or } n = 6$

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Question 15: In how many ways can five children stand in a queue?

We need to permute $\displaystyle 5$ children out of $\displaystyle 5$.

$\displaystyle \text{Therefore the required number of ways } = ^5P_5 = \frac{5!}{(5-5)!} = 5! = 120$

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Question 16: From among the $\displaystyle 36$ teachers in a school, one principal and one vice-principal are to be appointed. In how many ways can this be done?

We need to permute $\displaystyle 2$ teachers out of $\displaystyle 36$.

$\displaystyle \text{Therefore the required number of ways } = ^{36}P_2 = \frac{36!}{(36-2)!} = 36 \times 35 = 1260$

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Question 17: Four letters $\displaystyle \text{E, K, S}$ and $\displaystyle \text{ V }$, one in each, were purchased from a plastic warehouse. How many ordered pairs of letters, to be used as initials, can be formed from them?

We need to permute $\displaystyle 2$ letters out of $\displaystyle 4$.

$\displaystyle \text{Therefore the required number of ways } = ^4P_2 = \frac{4!}{(4-2)!} = 4 \times 3 = 12$

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Question 18: Four books, one each in Chemistry, Physics, Biology, and Mathematics, are to be arranged on a shelf. In how many ways can this be done?

We need to permute $\displaystyle 4$ books out of $\displaystyle 4$.

$\displaystyle \text{Therefore the required number of ways } = ^4P_4 = \frac{4!}{(4-4)!} = 4! = 24$

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Question 19: Find the number of different $\displaystyle 4-$letter words, with or without meanings, that can be formed from the letters of the word $\displaystyle \text{ 'NUMBER'}$.

We need to permute $\displaystyle 4$ letters out of $\displaystyle 6$.

$\displaystyle \text{Therefore the required number of ways } = ^6P_4 = \frac{6!}{(6-4)!} = 6 \times 5 \times 4 \times 3 = 360$

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Question 20: How many three-digit numbers are there, with distinct digits, with each digit odd?

The odd digits are $\displaystyle 1, 3, 5, 7, 9$

We need to permute $\displaystyle 3$ digits out of $\displaystyle 5$.

$\displaystyle \text{Therefore the required number of ways } = ^5P_3 = \frac{5!}{(5-3)!} = 5 \times 4 \times 3 = 60$

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Question 21: How many words, with or without meaning, can be formed by using all the letters of the word $\displaystyle \text{'DELHI'}$, using each letter exactly once?

Number of letters in $\displaystyle \text{DELHI}$ are $\displaystyle 5$

We need to permute $\displaystyle 5$ letters out of $\displaystyle 5$.

$\displaystyle \text{Therefore the required number of ways } = ^5P_5 = \frac{5!}{(5-5)!} = 5! = 120$

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Question 22: How, many words, with or without meaning, can be formed by using the letters of the word $\displaystyle \text{'TRIANGLE'}$?

Number of letters in $\displaystyle \text{TRIANGLE}$ are $\displaystyle 8$

We need to permute $\displaystyle 8$ letters out of $\displaystyle 8$.

$\displaystyle \text{Therefore the required number of ways } = ^8P_8 = \frac{8!}{(8-8)!} = 8!$

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Question 23: There are two works each of $\displaystyle 3$ volumes and two works each of $\displaystyle 2$ volumes; In how many ways can the $\displaystyle 10$ books be placed on a shelf so that the volumes of the same work are not separated?

There are 4 different types of works.

Number of ways in which these $\displaystyle 4$ works can be arranged $\displaystyle = ^4P_4 = 4!$

Within the works, the volumes can be arranged.

Therefore the total number of ways arrangement can be made $\displaystyle = 4! \times ( 3! \times 3!) \times ( 2! \times 2!) = 3456$.

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Question 24: There are $\displaystyle 6$ items in column $\displaystyle A$ and $\displaystyle 6$ items in column $\displaystyle B$. A student is asked to match each item in column $\displaystyle A$ with an item in column $\displaystyle B$. How many possible, correct or incorrect, answers are there to this question?

There are $\displaystyle 6$ items in column $\displaystyle A$ and $\displaystyle 6$ items in column $\displaystyle B$.

The first item from column $\displaystyle A$ can be matched with any of the $\displaystyle 6$ items from column $\displaystyle B$.

Similarly, once the first items from column $\displaystyle A$ has been matched, the 2nd item from the column $\displaystyle A$ can only be matched with $\displaystyle 5$ items of the column $\displaystyle B$. and so on….

Hence the number of ways the match can be done $\displaystyle = ^6P_6 = 6! = 720$

The other way to look at this is to keep column $\displaystyle A$ fixed and arrange 6 items from column $\displaystyle B$ which would be $\displaystyle ^6P_6 = 6! = 720$

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Question 25: How many three-digit numbers are there, with no digit repeated?

$\displaystyle \text{Total number of 3 digit number } = ^{10}P_3 = \frac{10!}{(10-3)!} = \frac{10!}{7!} = 10 \times 9 \times 8 = 720$

Total number of 3 digit numbers starting with $\displaystyle 0 = ^9P_2 = \frac{9!}{(9-2)!} = \frac{9!}{7!} = 9 \times 8 = 72$

Hence the three-digit numbers are there, with no digit repeated $\displaystyle = 720 - 72 = 648$

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Question 26: How many $\displaystyle 6-$digit telephone numbers can be constructed with digits $\displaystyle 0,1,2,3,4,5,6,7,8,9$ if each number starts with $\displaystyle 35$ and no digit appears more than once?

In total there are $\displaystyle 10$ digits to chose from.

However, the first two digits are already fixed to $\displaystyle 35$.

Therefore we can now chose $\displaystyle 4$ numbers from $\displaystyle 8$ in

$\displaystyle = ^8P_4 = \frac{8!}{4!} = 8 \times 7 \times 6 \times 5 = 1680$ ways.

Therefore $\displaystyle 1680$, $\displaystyle 6-$ digit telephone numbers can be constructed with digits $\displaystyle 0,1,2,3,4,5,6,7,8,9$ if each number starts with $\displaystyle 35$ and no digit appears more than once.

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Question 27: In how many ways can $\displaystyle 6$ boys and $\displaystyle 5$ girls be arranged for a group photograph if the girls are to sit on chairs in a row, and the boys are to stand in a row behind them?

Number of ways you can arrange the girls $\displaystyle = 5!$

Number of ways you can arrange the boys $\displaystyle = 6!$

Hence the total number of arrangements $\displaystyle = 5! \times 6! = 86400$

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Question 28: If $\displaystyle a$ denotes the number of permutations of $\displaystyle (x + 2)$ things taken all at a time, $\displaystyle b$ the number of permutations of $\displaystyle x$ things taken $\displaystyle 11$ at a time and $\displaystyle c$ the number of permutations of $\displaystyle x - 11$ things taken all at a time such that $\displaystyle a = 182 bc$, find the value of $\displaystyle x$.

$\displaystyle a = ^{x+2}P_{x+2} = \frac{(x+2)!}{x+2 - x - 2} = (x+2)!$

$\displaystyle b = ^{x}P_{11} = \frac{x!}{(x-11)!}$

$\displaystyle c = ^{x-11}P_{x-11 - x + 11} = \frac{(x-11)!}{x-11-x+11} = (x-11)!$

Given $\displaystyle a = 182 bc$

$\displaystyle \Rightarrow (x+2)! = 182 \cdot \frac{x!}{x-11} \cdot (x-11)!$

$\displaystyle \Rightarrow (x+2)! = 182 \cdot x!$

$\displaystyle \Rightarrow (x+2)(x+1) x! = 182 \cdot x!$

$\displaystyle \Rightarrow (x+2)(x+1) = 182$

$\displaystyle \Rightarrow (x+2)(x+1) = 14 \times 13$

Comparing LHS with RHS we get

$\displaystyle x+2 = 14 \Rightarrow x = 12$

$\displaystyle x+ 1 = 13 \Rightarrow x = 12$

$\displaystyle \therefore x = 12$

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Question 29: How many $\displaystyle 3-$digit numbers can be formed by using the digits $\displaystyle 1$ to $\displaystyle 9$ if no digit is repeated?

The total number of 3 digit number that can be formed $\displaystyle = ^9P_3 = \frac{9!}{6!} = 9 \times 8 \times 7 = 504$

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Question 30: How many 3-digit even numbers can be made using the digits $\displaystyle 1,2,3,4, 5,6,7$ if no digits is repeated?

The even digits are 2, 4, 6

$\displaystyle \text{ The number of 3-digit numbers when the last digit is } 2 = ^6P_2 = \frac{6!}{4!} = 6 \times 5 = 30$

$\displaystyle \text{ The number of 3-digit numbers when the last digit is } 4 = ^6P_2 = \frac{6!}{4!} = 6 \times 5 = 30$

$\displaystyle \text{ The number of 3-digit numbers when the last digit is } 6 = ^6P_2 = \frac{6!}{4!} = 6 \times 5 = 30$

Hence the total number of 3 digit even numbers that can be formed $\displaystyle = 30 + 30 + 30 = 90$

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Question 31: Find the numbers of $\displaystyle 4-$digit numbers that can be formed using the digits $\displaystyle 1, 2,3,4,5$ if no digit is repeated? How many of these will be even?

The number of 4 digit numbers that can be formed $\displaystyle = ^5P_4 = \frac{5!}{1!} = 5 \times 4 \times 3 \times 2 = 120$

The even digits are $\displaystyle 2, 4$

$\displaystyle \text{ The number of 4-digit numbers when the last digit is } 2 = ^4P_3 = \frac{4!}{1!} = 24$

$\displaystyle \text{ The number of 4-digit numbers when the last digit is } 4 = ^4P_3 = \frac{4!}{1!} = 24$

Therefore the total number of even numbers $\displaystyle = 24 + 24 = 48$

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Question 32: All the letters of the word $\displaystyle \text{'EAMCOT'}$ are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other.

Number of consonants $\displaystyle = 3 \text{( M, C, T)}$
Number of vowels $\displaystyle = 3 \text{( E, A, O)}$
$\displaystyle \text{Number of ways the vowels can be arranged } = ^4P_3 = \frac{4!}{1} = 24$
$\displaystyle \text{Number of ways consonants can be arranged } = 3P_3 = 3! = 6$
Hence the total number of arrangements in which no two vowels are adjacent to each other $\displaystyle = 24 \times 6 = 144$