JEE Main 2024 (Online) 8th April Evening ShiftQuestion 1
The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to :
Permutation and Combination
Solution
Concept. M, A, T can be taken 0, 1 or 2 times; H, E, I, C, S 0 or 1 time. Count selections of total size 5. Formulas used. (1) Number of selections = coefficient of $x^5$ in $(1+x+x^2)^3(1+x)^5$Step 1 - Expand. $(1+x+x^2)^3=1+3x+6x^2+7x^3+6x^4+3x^5+x^6$. $(1+x)^5$ coefficients: $1,5,10,10,5,1$. Step 2 - Coefficient of $x^5$. Pair $x^k$ from the first with $x^{5-k}$ from the second: $$1\cdot1+3\cdot5+6\cdot10+7\cdot10+6\cdot5+3\cdot1$$ Step 3 - Add. $1+15+60+70+30+3=179$. Answer: C