Question 1: Evaluate each of the following:

i) $^8P_3$         ii) $^{10}P_4$           iii) $^6P_6$          iv) $P(6, 4)$

Answer:

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

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

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

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

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 $P (5, r) = P (6, r - 1)$, find $r$.

Answer:

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

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

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

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

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

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

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

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

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

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

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

Hence $\Rightarrow r = 4$

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Question 3: If $5 P(4, n) = 6 \cdot P (5, n - 1)$, find $n$.

Answer:

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

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

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

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

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

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

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

Comparing LHS and RHS we get

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

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

Hence $n = 3$

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Question 4: If $P (n,5) =20 \cdot P(n, 3)$, find $n$.

Answer:

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

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

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

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

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

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

Comparing LHS and RHS we get

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

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

Hence $n = 8$

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

Answer:

$^nP_4 = 360$

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

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

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

Comparing LHS and RHS we get

$n = 6$

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

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

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

Hence $n = 6$

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

Answer:

$P (9, r) = 360$

$\Rightarrow ^9P_r = 360$

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

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

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

$\Rightarrow 9-r = 5$

$\Rightarrow r = 4$

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Question 7: If $P(11, r) = P (12, r - 1)$ find $r$.

Answer:

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

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

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

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

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

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

Comparing LHS and RHS we get

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

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

$\therefore r = 9$

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Question 8: If $P (n, 4) =12 \cdot P (n,2)$, find $n$.

Answer:

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

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

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

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

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

Comparing LHS and RHS we get

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

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

$\therefore n = 6$

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Question 9: If $P (n -1, 3) : P (n, 4) =1 :9$, find $n$.

Answer:

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

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

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

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

$\Rightarrow 9 = n - 1$

$\Rightarrow n = 10$

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Question 10: If $P (2n -1, n): P (2n +1, n -1) = 22:7$ find $n$.

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow \therefore n = 10$

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

Answer:

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

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

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

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

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

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

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

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

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

$\therefore n = 5$

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

Comparing LHS with RHS, we get

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

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

$\therefore r = 14$

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

Answer:

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

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

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

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

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

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

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

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

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

Answer:

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

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

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

Answer:

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

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 $\text{E, K, S}$ and $\text{ V }$, 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?

Answer:

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

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 in a shelf. In how many ways can this be done?

Answer:

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

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 $4-$letter words, with or without meanings, that can be formed from the letters of the word $\text{ 'NUMBER'}$.

Answer:

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

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?

Answer:

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

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

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 $\text{'DELHI'}$, using each letter exactly once?

Answer:

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

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

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 $\text{'TRIANGLE'}$ ?

Answer:

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

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

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

Answer:

There are 4 different types of works.

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

Within the works the volumes can be arranged.

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

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

Answer:

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

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

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

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

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

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

Answer:

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 $0 = ^9P_2 =$ $\frac{9!}{(9-2)!}$ $=$ $\frac{9!}{7!}$ $= 9 \times 8 = 72$

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

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

Answer:

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

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

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

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

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

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Question 27: In how many ways can $6$ boys and $5$ 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?

Answer:

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

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

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

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

Answer:

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

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

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

Given $a = 182 bc$

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

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

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

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

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

Comparing LHS with RHS we get

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

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

$\therefore x = 12$

$\\$

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

Answer:

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

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

Answer:

The even digits are 2, 4, 6

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

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

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 $= 30 + 30 + 30 = 90$

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

Answer:

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

The even digits are $2, 4$

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

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 $= 24 + 24 = 48$

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

Answer:

Number of consonants $= 3 \text{( M, C, T)}$

Number of vowels $= 3 \text{( E, A, O)}$

No two vowels can be together.

Number of ways the vowels can be arranged $= ^4P_3 =$ $\frac{4!}{1}$ $= 24$

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 $= 24 \times 6 = 144$