From the following words, select the one which cannot be formed using the letters of the word TOKENISM, assuming that no letter is used more times than it appears in the original word TOKENISM.

Introduction / Context:
This verbal reasoning problem asks you to determine which word cannot be formed from the letters of the base word TOKENISM. You are allowed to rearrange letters, but you must not introduce any new letters or use any existing letter more times than it occurs in the original word. This type of question is common in competitive exams and helps assess your attention to detail, especially your ability to track letter presence and frequency accurately under time pressure.

Given Data / Assumptions:

• Base word: TOKENISM.
• Available letters: T, O, K, E, N, I, S, M (each appears exactly once in the base word).
• Options: NAMES, EMITS, STONE, NOISE.
• Letters may be rearranged freely, but no new letters can be added and no letter can be repeated beyond its occurrence in TOKENISM.

Concept / Approach:
The standard strategy is to construct a mental or written set of letters that appear in the base word and then compare each option against this set. If every letter in an option appears in the base word with sufficient frequency, then that option can be formed. If an option uses a letter that is completely absent from the base word, or requires more copies of a letter than are available, that option is impossible to form and becomes the correct answer. Paying particular attention to vowels and unusual consonants often speeds up this process.

Step-by-Step Solution:
Step 1: List the letters of TOKENISM: T, O, K, E, N, I, S, M. Note that there is no letter A in this set. Step 2: Check EMITS. Its letters are E, M, I, T, S. All of these letters are present in TOKENISM, and each is used at most once, so EMITS can be formed. Step 3: Check STONE. Its letters are S, T, O, N, E. Every one of these letters appears in TOKENISM, so STONE can also be formed. Step 4: Check NOISE. Its letters are N, O, I, S, E. All of these are present in the base word, with no frequency conflicts, so NOISE can be formed as well. Step 5: Check NAMES. Its letters are N, A, M, E, S. Here the letter A appears, but A does not occur anywhere in TOKENISM. Therefore NAMES cannot be formed from TOKENISM.

Verification / Alternative check:
A quick verification technique is to write the base word and strike out the letters needed for each option. For EMITS, you can successively cross out E, M, I, T and S, all of which are present. Doing the same for STONE and NOISE shows that their letters exist in the base word. When you attempt this for NAMES, you immediately find that there is no A to cross out. Since even one missing letter is enough to make an option impossible, this confirms that NAMES is the only invalid formation.

Why Other Options Are Wrong:
EMITS is formed using E, M, I, T and S, all of which are directly available in TOKENISM. STONE uses S, T, O, N and E, again all present. NOISE uses N, O, I, S and E, which are also part of the letter set. Because each of these options can be assembled from the letters of TOKENISM without any conflict, none of them satisfies the requirement of being impossible to form, so they are not correct answers.

Common Pitfalls:
One common mistake is to focus on how familiar or meaningful an option word looks rather than checking the letters carefully. Another frequent error is mentally inserting letters that feel natural in English—like A in NAMES—without verifying that the base word actually provides them. To avoid these mistakes, always start by identifying the set of allowed letters from the base word and then systematically checking each option, paying extra attention to vowels and less common consonants.

Final Answer:
Since the base word TOKENISM does not contain the letter A, the word NAMES cannot be formed from its letters, while all other options can. Therefore NAMES is the correct answer.