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1

AIEEE 2007

MCQ (Single Correct Answer)
The set S = {1, 2, 3, ........., 12} is to be partitioned into three sets A, B, C of equal size. Thus $$A \cup B \cup C = S,\,A \cap B = B \cap C = A \cap C = \phi $$. The number of ways to partition S is
A
$${{12!} \over {{{(4!)}^3}}}\,\,$$
B
$${{12!} \over {{{(4!)}^4}}}\,\,$$
C
$${{12!} \over {3!\,\,{{(4!)}^3}}}$$
D
$${{12!} \over {3!\,\,{{(4!)}^4}}}$$

Explanation

The total number of ways is

$${}^{12}{C_4} \times {}^{12 - 4}{C_4} \times {}^{12 - 4 - 4}{C_4} = {}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4} = {{12!} \over {{{(4!)}^3}}}$$

2

AIEEE 2006

MCQ (Single Correct Answer)
At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is
A
5040
B
6210
C
385
D
1110

Explanation

A voter can give vote to either 1 candidate or 2 candidates or 3 candidates or 4 candidates.

Case 1 : When he give vote to only 1 candidate then no ways = $${}^{10}{C_1}$$

Case 2 : When he give vote to 2 candidates then no ways = $${}^{10}{C_2}$$

Case 3 : When he give vote to 3 candidates then no ways = $${}^{10}{C_3}$$

Case 4 : When he give vote to 4 candidates then no ways = $${}^{10}{C_4}$$
So, total no of ways he can give votes
= $${}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + {}^{10}{C_4}$$
= 385

Note : Here we use addition rule as he can vote any one of those four rules. Whenever there is "or" choices, we use addition rule.
3

AIEEE 2005

MCQ (Single Correct Answer)
If the letter of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number
A
601
B
600
C
603
D
602

Explanation

4

AIEEE 2004

MCQ (Single Correct Answer)
The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is
A
$${}^8{C_3}$$
B
21
C
$${3^8}$$
D
5

Explanation

To distribute n objects among p people where everyone should get atleast one object, then number of ways to distribute those n objects

= $${}^{n - 1}{C_{p - 1}}$$

For this question, n = 8 and p = 3

$$ \therefore $$ Number of ways = $${}^{8 - 1}{C_{3 - 1}}$$ = $${}^7{C_2}$$ = 21

Questions Asked from Permutations and Combinations

On those following papers in MCQ (Single Correct Answer)
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