Question

Consider the letters of the word MATHEMATICS. Set of repeating letters $=\{M,A,T\}$ and set of non-repeating letters $=\{H,E,I,C,S\}$

Column I		Column II
P	The number of words taking all letters of the given word such that atleast one repeating letter is at odd position is	1	$28\cdot 7!$
Q	The number of words formed taking all letters of the given word in which no two vowels are together is	2	$\frac{11!}{(2!{)}^3}$
R	The number of words formed taking all letters of the given word such that in each word both M's are together and both T's are together but both A's are not together is	3	$210\cdot 7!$
S	The number of words formed taking all letters of the given word such that relative order of vowels and consonants does not change is	4	$\frac{4!\cdot 7!}{(2!{)}^3}$

• A. P-2, Q-3, R-1, S-4
• B. P-3, Q-2, R-4, S-1
• C. P-3, Q-2, R-1, S-4
• D. P-2, Q-3, R-4, S-1

Answer 1

Answer: (A)

Even positions 2, 4, 6, 8, 10 (5)
Minimum one alphabets goes at odd possition
Hence, answer is total number of words formed without any conditions $\frac{11!}{(2!)(2!)(2!)}=\frac{11!}{(2!{)}^3}$ Ans.
(Q) 4 vowels (AA IE)
Rest $=7$ ( 8 gaps forms)
gap selection $={}^8{C}_4$
arrangement of vowels $=\frac{4!}{2!}$
arrangement of rest letters $=\frac{7!}{2!2!}$
$\Rightarrow {}^8{C}_4\times \frac{4!}{2!}\times \frac{7!}{2!2!}\Rightarrow 210\times 7!$ Ans.
(R) (MM) (TT) HEICS (7)
AA separate
8 gaps
${}^8{C}_2\times (1!)\times 7$ ! (for AA to be separated)
$\Rightarrow 28\times 7$ ! Ans.
(S) 4 vowels AAIE
7 consonants M M THC S
$\frac{4!}{2!}\times \frac{7!}{2!2!}$ (arrangement of vowels and consonants at their respective positions.)
$\Rightarrow \frac{4!\cdot 7!}{(2!{)}^3}$