Skip to main content
Back to PERMUTATIONS

JEE MATHEMATICS MCQ: PERMUTATIONS

1 Hard
1 of 1
Application +4 | -1
The number of ways of selecting nn objects out of 3n3 n objects, of which nn are alike and rest are different, is
View answer and explanation
Correct option: A
22n−1+(2n−1)!n!(n−1)!2^{2 n-1}+\frac{(2 n-1) !}{n !(n-1) !}
Option A (22n−1+(2n−1)!n!(n−1)!2^{2 n-1}+\frac{{(2 n-1) !}}{{n !(n-1) !}}) is correct because it represents the number of ways to select n objects out of 3n objects, where n are alike and the rest are different.

More Questions from PERMUTATIONS

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.