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$(a)$${46^{th}}$

$(b)$${59^{th}}$

$(c)$${52^{nd}}$

$(d)$${58^{th}}$

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Hint-Try and form all lexicographically smaller alphabets than word SMALL in the dictionary.

Here we have to tell the position of the word SMALL in the dictionary.

Now let’s fix A in the first position, then we get

A_ _ _ _,

Now we are left with 4 places in which we need to arrange 4 alphabets that is M, A, L, L. As L is repeating twice so the total ways of arranging 4 letters in 4 places in which two letters are repeating is \[\frac{{4!}}{{2!}} = \frac{{4 \times 3 \times 2}}{{2 \times 1}} = 12\]

Now let’s fix L in the first position, then we get

L_ _ _ _

Now we are left with 4 places in which we need to arrange 4 alphabets that is M, A, L, S. As no letter is repeated twice so the total ways of arranging 4 letters in 4 places is\[4! = 4 \times 3 \times 2 \times 1 = 24\]

Now let’s fix M in the first position, then we get

M_ _ _ _

Now we are left with 4 places in which we need to arrange 4 alphabets that is S, A, L, L. As L is repeating twice so the total ways of arranging 4 letters in 4 places in which two letters are repeating is \[\frac{{4!}}{{2!}} = \frac{{4 \times 3 \times 2}}{{2 \times 1}} = 12\]

Now let’s fix S in the first position and A in second position, then we get

S A _ _ _

Now we are left with 3 places in which we need to arrange 3 alphabets that is L, L, and M. As L is repeating twice so the total ways of arranging 3 letters in 3 places in which two letters are repeating is \[\frac{{3!}}{{2!}} = \frac{{3 \times 2}}{{2 \times 1}} = 3\]

Now let’s fix S in the first position and L in second position, then we get

S L _ _ _

Now we are left with 3 places in which we need to arrange 3 alphabets that is M, A, L. As no letter is repeated twice so the total ways of arranging 3 letters in 3 places is\[3! = 3 \times 2 \times 1 = 6\]

Now let’s fix S in the first position and M in second position, then if we arrange in alphabetical order the next word we get is SMALL only.

Hence the position of SMALL in dictionary is

Position $ = 12 + 24 + 12 + 3 + 6 + 1 = {58^{th}}$

Thus (d) is the right option

Note- The key concept while solving such problems is to make all the possible cases of words that are lexicographically smaller than the given word in the dictionary, this eventually gives us the position of the word in the dictionary.

Here we have to tell the position of the word SMALL in the dictionary.

Now let’s fix A in the first position, then we get

A_ _ _ _,

Now we are left with 4 places in which we need to arrange 4 alphabets that is M, A, L, L. As L is repeating twice so the total ways of arranging 4 letters in 4 places in which two letters are repeating is \[\frac{{4!}}{{2!}} = \frac{{4 \times 3 \times 2}}{{2 \times 1}} = 12\]

Now let’s fix L in the first position, then we get

L_ _ _ _

Now we are left with 4 places in which we need to arrange 4 alphabets that is M, A, L, S. As no letter is repeated twice so the total ways of arranging 4 letters in 4 places is\[4! = 4 \times 3 \times 2 \times 1 = 24\]

Now let’s fix M in the first position, then we get

M_ _ _ _

Now we are left with 4 places in which we need to arrange 4 alphabets that is S, A, L, L. As L is repeating twice so the total ways of arranging 4 letters in 4 places in which two letters are repeating is \[\frac{{4!}}{{2!}} = \frac{{4 \times 3 \times 2}}{{2 \times 1}} = 12\]

Now let’s fix S in the first position and A in second position, then we get

S A _ _ _

Now we are left with 3 places in which we need to arrange 3 alphabets that is L, L, and M. As L is repeating twice so the total ways of arranging 3 letters in 3 places in which two letters are repeating is \[\frac{{3!}}{{2!}} = \frac{{3 \times 2}}{{2 \times 1}} = 3\]

Now let’s fix S in the first position and L in second position, then we get

S L _ _ _

Now we are left with 3 places in which we need to arrange 3 alphabets that is M, A, L. As no letter is repeated twice so the total ways of arranging 3 letters in 3 places is\[3! = 3 \times 2 \times 1 = 6\]

Now let’s fix S in the first position and M in second position, then if we arrange in alphabetical order the next word we get is SMALL only.

Hence the position of SMALL in dictionary is

Position $ = 12 + 24 + 12 + 3 + 6 + 1 = {58^{th}}$

Thus (d) is the right option

Note- The key concept while solving such problems is to make all the possible cases of words that are lexicographically smaller than the given word in the dictionary, this eventually gives us the position of the word in the dictionary.

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