JEE PYQ: Permutation And Combination Question 21
Question 21 - 2020 (05 Sep Shift 2)
There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which the candidate can choose the questions, is:
(1) 3000
(2) 1500
(3) 2255
(4) 2250
Show Answer
Answer: (4)
Solution
Total $= 3 \times {}^5C_1 \times {}^5C_1 \times {}^5C_3 + 3 \times {}^5C_1 \times {}^5C_3 \times {}^5C_1 + 3 \times {}^5C_1 \times {}^5C_2 \times {}^5C_2 = 3 \times 5 \times 5 \times 10 + 3 \times 5 \times 10 \times 5 + … $ Wait – distributions $(1,1,3), (1,2,2), (1,3,1), (2,1,2), (2,2,1), (3,1,1)$: $3 \times {}^5C_1 {}^5C_1 {}^5C_3 + 3 \times {}^5C_1 {}^5C_2 {}^5C_2 = 3 \times 250 + 3 \times 500 = 750 + 1500 = 2250$.
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