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sritamasia
27 Nov 2021, 15:24
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
79% (01:02) correct
21%
(01:10)
wrong
based on 161
sessions
History
Date
Time
Result
Not Attempted Yet
There are 3 copies each of 4 different books. In how many ways can they be arranged on a shelf?
A. \(\frac{10!}{(3!)^{2}}\)
B. \(\frac{12!}{(3!)^{3}}\)
C. \(\frac{12!}{(3!)^{2}}\)
D. \(\frac{10!}{(3!)^{4}}\)
E.\(\frac{12!}{(3!)^{4}}\)
A. \(\frac{10!}{(3!)^{2}}\)
B. \(\frac{12!}{(3!)^{3}}\)
C. \(\frac{12!}{(3!)^{2}}\)
D. \(\frac{10!}{(3!)^{4}}\)
E.\(\frac{12!}{(3!)^{4}}\)
03 Dec 2021, 10:08
Kudos
Bookmarks
(n=12) total books
4 books with 3 copies each means that 12 should be divided by 3!x3!x3!x3! (number of repetition)
12! / (3!x3!x3!x3!) = 12!/3!^4
E should be the correct answer!
4 books with 3 copies each means that 12 should be divided by 3!x3!x3!x3! (number of repetition)
12! / (3!x3!x3!x3!) = 12!/3!^4
E should be the correct answer!
12 Dec 2021, 07:46
Kudos
Bookmarks
sritamasia
It's like no. of possibilities for arranging AAABBBCCCDDD.
Total \(12 \) objects however \(3\) of one kind are same .
Hence \(\frac{12!}{3!3!3!3!}\)
Ans = E
Hope it's clear.
