If the all S,P are separated in the word MISSISSIPPI then the total possible arrangements are $k\frac {10!}{4!.4!} $ then value of k is? Now all S,P are separated so we can have slash method. Thus they can be placed alternatively in $6! $ ways and rest 5 can be placed in $5! $ ways. Thus we equate to get $$6!.5!=k\frac {10!}{4!.4!} $$ to get $k=13.7$ while $k=4/3$ .Where is my mistake Thanks
-
Can a S and a P occur together ? – Shailesh Sep 06 '16 at 10:03
-
Yes they can as nothing is mentioned in the question – Archis Welankar Sep 06 '16 at 10:11
-
Your book answer is incorrect. $k = 1$ is correct as pointed out by the answer by trueblueanil. – Shailesh Sep 07 '16 at 05:54
1 Answers
Leaving out the $S's$ for the moment, there are $\frac{7!}{4!2!} = 105$ permutations of $PPMIIII$
Of these $\frac{6!}{4!} = 30$ will have the $P's$ together, thus $75$ will have the $P's$ apart.
For the together cases, we need to borrow an $S$ to separate them, e.g. $\boxed{PSP}MIIII$
The remaining $3\; S's$ can be inserted in the gaps between units (including ends) in $\binom73$ ways
For cases with $P's$ already separate, e.g. PMPIIII, the $4\; S's$ can be inserted similarly in $\binom84$ ways.
Thus permissible permutations $= 30\binom73 + 75\binom84 = 6300$
For better or worse, this give a value of $k=1$
Use of heavy artillery
To make assurance doubly sure, I have used Jair Taylor's formula to confirm the answer.
Define polynomials for $k\geq 1$ by $q_k(x) = \sum_{i=1}^k \frac{(-1)^{i-k}}{i!} {k-1 \choose i-1}x^i$.
e.g. for $k=2, q_2(x)$ works out to ${(x^2-2x)}/2!$
For the nonce, we shall take the $I's$ to be distinct, divide by $4!$ at the end of the calculations. So we have five distinct alphabets, one of length $2$ and one of length $4$
The number of permutations will be given by
$$\int_0^\infty \prod_j q_{k_j}(x)\, e^{-x}\,dx.$$
Wolframalpha gives the answer as $151,200$
Dividing by $4!$ to take care of the $4 I's$, we again get the answer of $6300$
- 209,954
- 41,295
-
Your argument is flawless - not checked the 'heavy artillery' one. Also one Aditya Raut confirmed 6300 with Python code. +1 for all the efforts. – Shailesh Sep 07 '16 at 05:52
-
Im just a beginner so I didnt see you computer techniques but the first part os awesom thanks – Archis Welankar Sep 07 '16 at 10:49
-
-