12 votes 12 votes The number of ways in which $5\; A's, 5\; B's$ and $5\; C's$ can be arranged in a row is: $15!/(5!)^{3}$ $15!$ $\left(\frac{15}{5}\right)$ $15!(5!3!)$. Combinatory gate1990 normal combinatory + – makhdoom ghaya asked Nov 22, 2016 • retagged Jul 19, 2023 by shadymademe makhdoom ghaya 3.1k views answer comment Share Follow See 1 comment See all 1 1 comment reply sauravgahlawat commented Dec 12, 2021 reply Follow Share why there’s a multi-select tag? 1 votes 1 votes Please log in or register to add a comment.
Best answer 11 votes 11 votes (A) Use permutation with repetitions formula as we have to arrange $15$ elements where $5$ each are identical. NRN answered Jan 2, 2017 • edited Jun 6, 2018 by Arjun NRN comment Share Follow See all 3 Comments See all 3 3 Comments reply Lakshman Bhaiya commented Feb 9, 2019 reply Follow Share https://brilliant.org/wiki/permutations-with-repetition/ 5 votes 5 votes KUSHAGRA गुप्ता commented Jan 14, 2020 reply Follow Share Permutation with constrained repetitions: Number of permutations of n-objects in which q1 are of same type q2 are of same type . . $=\dfrac{n!}{q1!\ q2!\ q3!...qn!}$ 1 votes 1 votes halfcodeblood commented Apr 10 reply Follow Share Total 15 and each one is repeating 5 times So answer will 15!/5!5!5! or 15!/(5!)^3 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes The longer way Option $\large A$ subbus answered Jul 17, 2021 • reshown Jul 24, 2021 by subbus subbus comment Share Follow See all 0 reply Please log in or register to add a comment.