Consider all possible permutations of the letters of the word INDIANOIL. Match the entries in Column I with the correctly related quantum number(s) in Column II
Column-I Column-II
A) The number of permutations containing the word ‘INDIA’ is p) 5!
B) The number of permutations in which the letter ‘I’ 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 5 positions is r) 7 × 5!
D) The number of permutations in which the letters ‘A, I, O’ occur only in odd positions is s) 21 × 5!

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A-p; B-s; C-r; D-q

A-s; B-p; C-r; D-q

A-s; B-p; C-q; D-q

A-p; B-s; C-q; D-q

answer is A.

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Detailed Solution

INDIA

$NOIL \to I 3, N - 2, D - 1, O - 1, A - 1, L - 1$

$A) (INDIA), N,O, I,L \Rightarrow 5!$

$B) \Rightarrow \frac{7!}{2!} = 21 \times 5!$

$C) N, D, L, N at first four positions and rest five i.e. E, E, E, A, O at last five posi tions. - \dots \frac{5!}{3!} \times \frac{4!}{2!} = 2 \times 5!$

$D) \bar{1} \bar{2} \bar{3} \bar{4} \bar{5} \bar{6} \bar{7} \bar{8} \bar{9}$

$There are 5 odd places 1, 3, 5, 7, 9$

$A, E, E, E, O can be arranged at 1^{st}, 3^{rd}, 5^{th}, 7^{th} and 9^{th} positions and rest N, D, N, L at rest 4 positions. This can be done in \Rightarrow \frac{5!}{3!} \times \frac{4!}{2!} = 2 \times 5!$

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Column-I	Column-II
A) The number of permutations containing the word ‘INDIA’ is	p) 5!
B) The number of permutations in which the letter ‘I’ 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 5 positions is	r) 7 × 5!
D) The number of permutations in which the letters ‘A, I, O’ occur only in odd positions is	s) 21 × 5!