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 1^st ,3^rd ,5^th ,7^th and 9^th => 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)

Answer is:

(a) -> (p)         (b) -> (s)          (c) -> (q)          (d) -> (q)