Here is the solved problems from previous year IIT-JEE papers on Combinatorics

**Q. [IIT-JEE 2007]**The letters of the word COCHIN are permitted and all the permutations are arranged in an English dictionary. The number of words that appear before the word COCHIN, is

(a) 360
(b) 192 (c) 96 (d) 48

__Solution :__(c) The given word COCHIN has 6 letters with C repeated twice. The word begin with CX where X can be any one of the letters C,H,I,N.

Fixing one C in the first place, the second place
can be filled in 4 ways by any of the letters C,H,I,N and remaining four places
can be filled in 4! Ways. Thus, number of words before COCHIN is = 4×(4!).

**Q. [IIT-JEE 2001]**An n-digit number is a positive number with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this is possible is

(a) 5
(b) 7 (c) 8 (d) 9

__Solution :__(b) Given digits are 2,5 and 7 and n-digit number is to be formed using these digits. Out of n places, each place can be filled in 3 ways. Thus total number of ways = 3

^{n}. Now, 3

^{n}≥ 900 = 3

^{2}×100 ⇒ n – 2 ≥ 5 .

Least
value of n is 7.

**Q. [IIT-JEE 1998]**Number of divisors of the form 4n+2 (n ≥ 0) of integer 240 is :

(a) 4 (b) 8
(c) 10 (d) 3

__Solution :__(a) The prime factorisation of 240 is 240 = 2

^{4}×3

^{1}×5

^{1}

Total
number of divisors of 240 is = (4+1)(1+1)(1+1) = 20. Divisors can be 2,3,5 or
their products. Now divisors of the form 4n + 2 are 2,6,10,30.