# some counting problems

(1) The no. of words that can be made using the letter of the word $\bf{PARAMETER}$ so that no vowel in between two consonants.

(2) A password, which consists of the digits from $0$ to $9,$ and is of length $4$ that uses exactly $2$ different digits. How many different passwords are possible.

(3) Total no. of all possible arrangement of the word $\bf{MATHEMATICS}$ in which no two $\bf{M}$

and no two $\bf{A}$ occur together, is

$\bf{My\;\; Try}::$ for (1) Here vowels $\bf{A,A,E,E}$ and consonants $\bf{P,R,R,M,T}$

Now we will take all vowels as one word $\bf{\boxed{A,A,E,E},P,R,R,M,T}$

So Total arrangement in which no vowel in between two consonants, is

$\displaystyle = \frac{6!}{2!}\times \frac{4!}{2!\times 2!} = 6\times 5 \times 4 \times 3 \times 2 \times 3 = 2160$

for (II) one

Here we have to form $4$ Digit password in which exactly $2$ Digit are distinct

So we have form two cases

$\bf{case\; (I)::}$ If one distinct digit and other three are same like $abbb$ or $aaab$

Which can be done in $\displaystyle = \binom{9}{1}\times \binom{8}{1}\times \frac{4!}{3!} = 72\times 4$

$\bf{case\; (II)::}$ If two are of one type digit and other two are of same type $aabb$

Which can be done in $\displaystyle = \binom{9}{1}\times \binom{8}{1}\times \frac{4!}{2!\times 2!} = 72 \times 6$

So Total is $= 72 \times (4+6) = 720$

Now I did not understand how can i solved third one.

also plz explain me is I am right or not for two above questions.

Thanks

-

Consider the cases when $AA$ and $MM$ are grouped together. Now there are 9 "objects" so to speak. It follows that the number of permutations is $\frac{9!}{2!} = 181.440$. Now when just $AA$ or $MM$ is grouped together we have $\frac{10!}{(2!)^{2}} = 907.200$. Now, the total number of permutations where either $AA$ or $MM$ are grouped together is $907.200\cdot 2 - 181440 = 1.632.960$. Now, note that the total number of permutations is: $\frac{11!}{(2!)^{3}} = 4.989.600$. So our final answer is $4.989.600-1.632.960=3.356.640$ ways.