Find the number of permutations of the letters of the words
$'DADDY DID A DEADLY DEED'$

\frac{19!}{9! \times 3! \times 2! \times 3!}

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\frac{19!}{10! \times 2! \times 2! \times 3!}

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\frac{19!}{9! \times 3! \times 4! \times 3!}

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\frac{19!}{8! \times 3! \times 2! \times 3!}

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Solution

The correct option is D $\frac{19!}{9! \times 3! \times 2! \times 3!}$
Here, total letters $19$ , in which $9 D^{'} s$ , $3 A^{'} s$ , $2 Y^{'} s$ and $3 E^{'} s$ and rest occur only once.
$∴$ Required number of permutations = $\frac{19!}{9! \times 3! \times 2! \times 3!}$

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Similar questions

Add $t - t^{2} - 14;$ $15 t^{3} + 13 + 9 t - 8 t^{2}$ ; $12 t^{2} - 19 - 24 t$ and $4 t - 9 t^{2} + 19 t^{3}$ . [4 MARKS]

Add
$t - t^{2} - 14$ , $15 t^{3} + 13 + 9 t - 8 t^{2}$ , $12 t^{2} - 19 - 24 t$ and $4 t - 9 t^{2} + 19 t^{3}$ .

If $t = - 1$ find the value of the expression. [4 MARKS]

Find the sum:

(i) (ii) (iii)

(iv) (v) (vi)

(vii)

Q. Divide

10 x^{3} + 19 x^{2} + 12 x - 9

2 x - 3

Q. When simplified

{(- \frac{1}{27})}^{- 2 / 3}

is

(a) 9

(b) −9

(c)

\frac{1}{9}

(d)

- \frac{1}{9}

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Find the number of permutations of the letters of the words 'DADDY DID A DEADLY DEED'

The correct option is D 19!9!×3!×2!×3!Here, total letters 19, in which 9D′s, 3A′s, 2Y′s and 3E′s and rest occur only once.∴ Required number of permutations = 19!9!×3!×2!×3!

Find the number of permutations of the letters of the words
$'DADDY DID A DEADLY DEED'$

The correct option is D $\frac{19!}{9! \times 3! \times 2! \times 3!}$
Here, total letters $19$ , in which $9 D^{'} s$ , $3 A^{'} s$ , $2 Y^{'} s$ and $3 E^{'} s$ and rest occur only once.
$∴$ Required number of permutations = $\frac{19!}{9! \times 3! \times 2! \times 3!}$