## Tuesday, 8 November 2011

### P&C-9 (Matching) (IIT-JEE 2008)

Consider all possible permutations of the letters of the word ENDEANOEL. Match the statements/expressions in Column-I with the statements/expressions in Column-II:
 Column-I Column-II (a)   The number of permutations containing the word ENDEA is (p) 5! (b) The number of permutations in which the letter E occurs in the first and the last positions is (q) 2*5! (C) The number of permutations in which none of the letters D,L,N occurs in the last five positions is (r) 7*5! (d) The number of permutations in which the letters A,E,O occur only in odd positions is (s) 21*5!

Solution follows here:
Solution:
“ENDEANOEL”
Number of letters:    E’s-3    N’s-2   D’s-1   A’s-1   O’s-1   L’s-1
Total 9 letters
(a)   The number of permutations containing the word ENDEA is:
Considering “ENDEA” as an entity, the remaining letters are – N,O,E,L
∴ all entities are: ENDEA,N,O,E,L
=>Total number of entities =5
=> As there are no repeat entities, total number of permutations = 5!
(a) -> (p)
(b) The number of permutations in which the letter E occurs in the first and the last positions is:
E _ _ _ _ _ _ _ E
If E occupy first and last positions, the remaining 7 positions can be occupied by the remaining letters, N,D,E,A,N,O,L in 7!/2! = 7*6*5!/2 = 21*5! Ways
(2 in the denominator because- ‘N’ occurs 2 times)
(b) -> (s)
(C) The number of permutations in which none of the letters D,L,N occurs in the last five positions is:
D’s-1   N’s-2   L’s-1 => These are 4 letters => given that these none of these 4 letters occurs in last five positions
=> these 4 letters occur in the first four positions in 4!/2! Ways and remaining 5 letters (E’s-3      A’s-1   O’s-1) occur in the last five positions in 5!/3! Ways
=> total number of permutations =   (4!/2!)(5!/3!) = 2*5!
(c) -> (q)
(d) The number of permutations in which the letters A,E,O occur only in odd positions is:
Odd positions are 1st ,3rd ,5th ,7th and 9th => total 5 positions
E’s-3    A’s-1   O’s-1 => total 5 letters
=> These 5 letters occur in the 5 odd positions in 5!/3! Ways and the remaining 4 letters (N’s-2   D’s-1   L’s-1) occur in the remaining 4 positions in 4!/2! Ways
=> total number of permutations =   (5!/3!) (4!/2!) = 2*5!
(d) -> (q)