Consider all possible permutations of
the letters of the word ENDEANOEL. Match the statements/expressions in ColumnI
with the statements/expressions in ColumnII:
Solution:
ColumnI

ColumnII

(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:
“ENDEANOEL”
Number of letters: E’s3 N’s2 D’s1 A’s1 O’s1 L’s1
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’s1 N’s2 L’s1 => 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’s3 A’s1 O’s1)
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’s3 A’s1 O’s1 => total 5 letters
=> These 5 letters occur in the 5
odd positions in 5!/3! Ways and the remaining 4 letters (N’s2 D’s1 L’s1)
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)
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