Question

A die is rolled 20 times. Given that three of the rolls came up number 1, five came up number 2, four came up number 3, two came up number 4, three came up number 5, and three came up number 6, how many different arrangements of the outcomes are there?

Answer 1

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!T\!otal\;number\;o\!f\;di\!f\!\!f\!erent\;\,outcomes =\binom{20}{3}\binom{17}{5}\binom{12}{4}\binom{8}{2}\binom{6}{3}\binom{3}{3}=\frac{20!}{3!\,5!\,4!\,2!\,3!\,3!}$

LOGIC 1 :-

First among 20 rolls, 3 rolls are chosen which come up with number 1. Then among remaining 17 rolls, 5 rolls are chosen which come up with number 2. Then among remaining 12 rolls, 4 rolls are chosen which come up with number 3. Then among remaining 8 rolls, 2 rolls are chosen which come up with number 4. Then among remaining 6 rolls, 3 rolls are chosen which come up with number 5. Then the remaining 3 rolls come up with the number 6.

So number of different outcomes

$=\binom{20}{3}\binom{17}{5}\binom{12}{4}\binom{8}{2}\binom{6}{3}\binom{3}{3}$

LOGIC 2 :-

Number of permutations of k distinct items where i-th item is repeated r_i times (taking all at a time) such that n=r₁+r₂+....+r_k

$=\frac{n!}{r_{1}!\,r_{2}!\,....\,r_{k}!}$

Here; n=20, k=6, r₁=3, r₂=5, r₃=4, r₄=2, r₅=3, r₆=3

So number of different outcomes

$=\frac{20!}{3!\,5!\,4!\,2!\,3!\,3!}$