QBManager

Permutation and Combination

119233 There are 3 copies each of 4 different books. The number of ways they can be arranged in a shelf is

1 369600

2 400400

3 420600

4 440720

Explanation:

A Given that, there are 3 copies each of 4
different books.
Total number of copies $= 12$
The desired number is $\frac{12!}{(3!)^{4}} = \frac{479001600}{1296} = 369600$

COMEDK-2016

Permutation and Combination

119234 A library has a copies of one book, b copies of of two books, c copies of three books and single copy of $d$ books. The total number of ways in which these books can be arranged in a shelf, is

1

\frac{(a + b + c + d)!}{a! b! c!}

2

\frac{(a + 2 b + 3 c + d)!}{a! (b!)^{2} (c!)^{3}}

3

\frac{(a + 2 b + 3 c + d)!}{a! b! c!}

4 None of these

Explanation:

B Here total number of books $= a + 2 b + 3 c + d$
Number of ways they can arranged $= (a + 2 b + 3 c + d)$ !
$∴$ Total number of arrangements
$= \frac{(a + 2 b + 3 c + d)!}{a! (b!)^{2} (c!)^{3}}$

COMEDK-2018

Permutation and Combination

119236 Six identical coins are arranged in a row. The number of ways in which the number of tails is equal to the number of heads is

1 20

2 9

3 120

4 40

Explanation:

A Here, the required number of ways is the number of ways in which the letters HHHTTT can be arranged. ( $∵ H =$ heads, $T =$ tails $)$
Hence, they can be all together arranged in 6 ! ways but, we have to account for 3 repetitive heads and tails.
Thus, the required permutation is
$= \frac{6!}{3! 3!} =^{6} C_{3} = \frac{6 \times 5 \times 4}{3!} = 20$

VITEEE-2019

Permutation and Combination

119240 There are 7 identical white balls and 3 identical black balls. The number of distinguishable arrangements in a row of all the balls, so that no two black balls are adjacent is

1 120

2

89 . (8!)

3 56

4

42 \times 5^{4}

Explanation:

C : First arrange 7 identical white ball in row is
$\frac{7!}{7!} = 1$ way.
Now for black balls we have 8 places from 8 places we can choose any of 3 places $=^{8} C_{3}$ and we can arrange them in 3 ! ways as all the 3 balls are identical so it should be $\frac{3!}{3!}$ hence, solution should be $\frac{7!}{7!} \times^{8} C_{3} \times \frac{3!}{3!} = 56$

AP EAMCET-19.08.2021

Permutation and Combination

119233 There are 3 copies each of 4 different books. The number of ways they can be arranged in a shelf is

1 369600

2 400400

3 420600

4 440720

Explanation:

A Given that, there are 3 copies each of 4
different books.
Total number of copies $= 12$
The desired number is $\frac{12!}{(3!)^{4}} = \frac{479001600}{1296} = 369600$

COMEDK-2016

Permutation and Combination

119234 A library has a copies of one book, b copies of of two books, c copies of three books and single copy of $d$ books. The total number of ways in which these books can be arranged in a shelf, is

1

\frac{(a + b + c + d)!}{a! b! c!}

2

\frac{(a + 2 b + 3 c + d)!}{a! (b!)^{2} (c!)^{3}}

3

\frac{(a + 2 b + 3 c + d)!}{a! b! c!}

4 None of these

Explanation:

B Here total number of books $= a + 2 b + 3 c + d$
Number of ways they can arranged $= (a + 2 b + 3 c + d)$ !
$∴$ Total number of arrangements
$= \frac{(a + 2 b + 3 c + d)!}{a! (b!)^{2} (c!)^{3}}$

COMEDK-2018

Permutation and Combination

119236 Six identical coins are arranged in a row. The number of ways in which the number of tails is equal to the number of heads is

1 20

2 9

3 120

4 40

Explanation:

A Here, the required number of ways is the number of ways in which the letters HHHTTT can be arranged. ( $∵ H =$ heads, $T =$ tails $)$
Hence, they can be all together arranged in 6 ! ways but, we have to account for 3 repetitive heads and tails.
Thus, the required permutation is
$= \frac{6!}{3! 3!} =^{6} C_{3} = \frac{6 \times 5 \times 4}{3!} = 20$

VITEEE-2019

Permutation and Combination

119240 There are 7 identical white balls and 3 identical black balls. The number of distinguishable arrangements in a row of all the balls, so that no two black balls are adjacent is

1 120

2

89 . (8!)

3 56

4

42 \times 5^{4}

Explanation:

C : First arrange 7 identical white ball in row is
$\frac{7!}{7!} = 1$ way.
Now for black balls we have 8 places from 8 places we can choose any of 3 places $=^{8} C_{3}$ and we can arrange them in 3 ! ways as all the 3 balls are identical so it should be $\frac{3!}{3!}$ hence, solution should be $\frac{7!}{7!} \times^{8} C_{3} \times \frac{3!}{3!} = 56$

AP EAMCET-19.08.2021

Permutation and Combination

119233 There are 3 copies each of 4 different books. The number of ways they can be arranged in a shelf is

1 369600

2 400400

3 420600

4 440720

Explanation:

A Given that, there are 3 copies each of 4
different books.
Total number of copies $= 12$
The desired number is $\frac{12!}{(3!)^{4}} = \frac{479001600}{1296} = 369600$

COMEDK-2016

Permutation and Combination

119234 A library has a copies of one book, b copies of of two books, c copies of three books and single copy of $d$ books. The total number of ways in which these books can be arranged in a shelf, is

1

\frac{(a + b + c + d)!}{a! b! c!}

2

\frac{(a + 2 b + 3 c + d)!}{a! (b!)^{2} (c!)^{3}}

3

\frac{(a + 2 b + 3 c + d)!}{a! b! c!}

4 None of these

Explanation:

B Here total number of books $= a + 2 b + 3 c + d$
Number of ways they can arranged $= (a + 2 b + 3 c + d)$ !
$∴$ Total number of arrangements
$= \frac{(a + 2 b + 3 c + d)!}{a! (b!)^{2} (c!)^{3}}$

COMEDK-2018

Permutation and Combination

119236 Six identical coins are arranged in a row. The number of ways in which the number of tails is equal to the number of heads is

1 20

2 9

3 120

4 40

Explanation:

A Here, the required number of ways is the number of ways in which the letters HHHTTT can be arranged. ( $∵ H =$ heads, $T =$ tails $)$
Hence, they can be all together arranged in 6 ! ways but, we have to account for 3 repetitive heads and tails.
Thus, the required permutation is
$= \frac{6!}{3! 3!} =^{6} C_{3} = \frac{6 \times 5 \times 4}{3!} = 20$

VITEEE-2019

Permutation and Combination

119240 There are 7 identical white balls and 3 identical black balls. The number of distinguishable arrangements in a row of all the balls, so that no two black balls are adjacent is

1 120

2

89 . (8!)

3 56

4

42 \times 5^{4}

Explanation:

C : First arrange 7 identical white ball in row is
$\frac{7!}{7!} = 1$ way.
Now for black balls we have 8 places from 8 places we can choose any of 3 places $=^{8} C_{3}$ and we can arrange them in 3 ! ways as all the 3 balls are identical so it should be $\frac{3!}{3!}$ hence, solution should be $\frac{7!}{7!} \times^{8} C_{3} \times \frac{3!}{3!} = 56$

AP EAMCET-19.08.2021

Permutation and Combination

119233 There are 3 copies each of 4 different books. The number of ways they can be arranged in a shelf is

1 369600

2 400400

3 420600

4 440720

Explanation:

A Given that, there are 3 copies each of 4
different books.
Total number of copies $= 12$
The desired number is $\frac{12!}{(3!)^{4}} = \frac{479001600}{1296} = 369600$

COMEDK-2016

Permutation and Combination

119234 A library has a copies of one book, b copies of of two books, c copies of three books and single copy of $d$ books. The total number of ways in which these books can be arranged in a shelf, is

1

\frac{(a + b + c + d)!}{a! b! c!}

2

\frac{(a + 2 b + 3 c + d)!}{a! (b!)^{2} (c!)^{3}}

3

\frac{(a + 2 b + 3 c + d)!}{a! b! c!}

4 None of these

Explanation:

B Here total number of books $= a + 2 b + 3 c + d$
Number of ways they can arranged $= (a + 2 b + 3 c + d)$ !
$∴$ Total number of arrangements
$= \frac{(a + 2 b + 3 c + d)!}{a! (b!)^{2} (c!)^{3}}$

COMEDK-2018

Permutation and Combination

119236 Six identical coins are arranged in a row. The number of ways in which the number of tails is equal to the number of heads is

1 20

2 9

3 120

4 40

Explanation:

A Here, the required number of ways is the number of ways in which the letters HHHTTT can be arranged. ( $∵ H =$ heads, $T =$ tails $)$
Hence, they can be all together arranged in 6 ! ways but, we have to account for 3 repetitive heads and tails.
Thus, the required permutation is
$= \frac{6!}{3! 3!} =^{6} C_{3} = \frac{6 \times 5 \times 4}{3!} = 20$

VITEEE-2019

Permutation and Combination

119240 There are 7 identical white balls and 3 identical black balls. The number of distinguishable arrangements in a row of all the balls, so that no two black balls are adjacent is

1 120

2

89 . (8!)

3 56

4

42 \times 5^{4}

Explanation:

C : First arrange 7 identical white ball in row is
$\frac{7!}{7!} = 1$ way.
Now for black balls we have 8 places from 8 places we can choose any of 3 places $=^{8} C_{3}$ and we can arrange them in 3 ! ways as all the 3 balls are identical so it should be $\frac{3!}{3!}$ hence, solution should be $\frac{7!}{7!} \times^{8} C_{3} \times \frac{3!}{3!} = 56$

AP EAMCET-19.08.2021

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