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JEE Main Mock Test Series 2020 & Previous Year Papers

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Illustration 1: The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear in before the word Cochin is
1. 360
2. 192
3. 96
4. 48 (2007)
Solution: We first arrange the letters of the word COCHIN as they would appear in an English Dictionary.
We get CCHINO.
Now, consider the words starting from C. There are 5! such words. Number of words with the two Cs occupying first and second place = 4!

Illustration 2: A committee of 12 is to be formed from 9 women and 8 men. In how many ways can this be done if at least five women have to be included in a committee? In how many of these committees
(a) The women are in majority?
(b) The men are in majority?        (1994)
Solution: Given that there are 9 women and 8 men, a committee of 12 is to be formed in such a way that it includes at least five women.
We first list the possible number of ways in which this can actually happen
= (5 women and 7 men) + (6 women and 6 men) + (7 women and 5 men) + (8 women and 4 men) + (9 women and 3 men)
Hence, total number of ways of forming the committee is given by
= (9C5.8C7) + (9C6.8C6) + (9C7.8C5) + (9C8.8C4) + (9C9.8C3)
= 1008 + 2352 + 2016 + 630 + 56
= 6062 ways

Illustration 3: 7 relatives of a man comprise 4 ladies and 3 gentlemen; his wife also has 7 relatives, 3 of which are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are three of man’s relatives and 3 of the wife’s relatives? (1985)
Solution: In such a situation, any of the following four cases can arise:
Case 1: A man invites 3 ladies and the woman invites 3 gentlemen:
Hence, the number of ways in this case is = 4C3.4C3 = 16
Case 2: The man invites 2 ladies, 1 gentleman and the woman invites 2 gentlemen, 1 lady
Hence, the number of ways in this case = (4C2.3C1). (3C1.4C2) = 324
Case 3: A man invites 1 lady, 2 gentlemen and the woman invites 2 ladies, 1 gentleman
Hence, the number of ways in this case = (4C1.3C2). (3C2.4C1) = 144
Case 4: A man invites 3 gentlemen and woman invites 3 ladies
Number of ways = 3C3. 3C3 = 1
Hence, the total required number of ways = 16 + 324 + 144 + 1 = 485.

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