The number of different arrangements of the letters of the word LETTER is
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Answer: (a) \(180\)
Six letters with E twice and T twice: \(\dfrac{6!}{2!\,2!} = \dfrac{720}{4} = 180\).
Permutations, combinations and the binomial theorem: JEE Main-style questions
Original JEE Main-style questions on permutations, combinations and the binomial theorem, with a worked solution and the usual trap for each. Tap an option, or type a numerical answer, to check it.
- 6 questions
- 3 free
- 4 multiple choice, 2 numerical value
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Learn the chapters first: Class 11, Permutations and Combinations · Class 11, Binomial Theorem.
Questions
The coefficient of \(x^4\) in \((1 + 2x)^6\) is
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Answer: (b) \(240\)
General term \(\binom{6}{r}(2x)^r\). For \(r = 4\): \(\binom{6}{4} \cdot 2^4 = 15 \times 16 = 240\).
Trap: Forgetting to raise the 2 to the fourth power gives \(\binom64 = 15\) or \(2 \times 15\).
Find the term independent of \(x\) in the expansion of \(\left(x^3 + \dfrac{2}{x}\right)^8\).
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Answer: 1792
\(T_{r+1} = \binom{8}{r}(x^3)^{8 - r}\left(\dfrac{2}{x}\right)^r = \binom{8}{r}2^r x^{24 - 4r}\). Independent of \(x\) when \(r = 6\).
Term \(= \binom{8}{6} \cdot 2^6 = 28 \times 64 = 1792\).
Trap: Forgetting the \(2^r\) from \(\left(\tfrac{2}{x}\right)^r\) gives 28.
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