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1

AIEEE 2011

MCQ (Single Correct Answer)

Statement - 1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is emply is $${}^9{C_3}$$.
Statement - 2: The number of ways of choosing any 3 places from 9 different places is $${}^9{C_3}$$.
A
Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1.
B
Statement - 1 is true, Statement - 2 is false.
C
Statement - 1 is false, Statement - 2 is true.
D
Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1.

Explanation

Let XA, XB, XC and XD represent number of balls present in box A, B, C and D respectively.

As no box can be empty so,

XA $$\ge$$ 1, XB $$\ge$$ 1, XC $$\ge$$ 1 and XD $$\ge$$ 1

$$\Rightarrow$$ XA $$-$$ 1 $$\ge$$ 0, $$\Rightarrow$$ XB $$-$$ 1 $$\ge$$ 0, $$\Rightarrow$$ XC $$-$$ 1 $$\ge$$ 0 and $$\Rightarrow$$ XD $$-$$ 1 $$\ge$$ 0

tA $$\ge$$ 0, tB $$\ge$$ 0, tC $$\ge$$ 0 and tD $$\ge$$ 0

According to the question,

XA + XB + XC + XD = 10

$$\Rightarrow$$ (XA $$-$$ 1) + (XB $$-$$ 1) + (XC $$-$$ 1) + (XD $$-$$ 1) = 6

$$\Rightarrow$$ tA + tB + tC + tD = 6

Now question becomes, box A, B, C, and D can have none or one or more balls and total balls are 6

From formula we know, n things can be distributed among r people in $${}^{n + r - 1}{C_{r - 1}}$$ ways where each people can have either 0 or more things.

$$\therefore$$ 6 balls can be distributed among 4 boxes in $${}^{6 + 4 - 1}{C_{4 - 1}} = {}^9{C_3}$$ ways where each box can have either 0 or more balls.

Therefore, Statement 1 is correct. The number of ways of choosing any 3 places from 9 different places is $${}^9{C_3}$$ ways. But Statement - 2 is not the correct explanation of Statement - 1.

2

AIEEE 2011

MCQ (Single Correct Answer)
These are 10 points in a plane, out of these 6 are collinear, if N is the number of triangles formed by joining these points. then:
A
$$N \le 100$$
B
$$100 < N \le 140$$
C
$$140 < N \le 190\,$$
D
$$N > 190$$

Explanation

We need 3 points to create a triangle. With 10 points number of triangle possible $${}^{10}{C_3}$$

Here 6 points are on the same line so we can't make any triangle with those 6 points.

So subtract $${}^{6}{C_3}$$.

$$\therefore$$ $$N = {}^{10}{C_3} - {}^6{C_3}$$

$$ = {{10\,.\,9\,.\,8} \over {1\,.\,2\,.\,3}} - {{6\,.\,5\,.\,4} \over {1\,.\,2\,.\,3}}$$

$$ = 120 - 20$$

$$ = 100$$

3

AIEEE 2010

MCQ (Single Correct Answer)
There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is
A
36
B
66
C
108
D
3

Explanation

Thus number of ways $$ = ({}^3{C_2}) \times ({}^9{C_2}) = 3 \times {{9 \times 8} \over 2} = 108$$

4

AIEEE 2009

MCQ (Single Correct Answer)
From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangement is :
A
at least 500 but less than 750
B
at least 750 but less than 1000
C
at least 1000
D
less than 500

Explanation

From 6 different novels 4 novels can be chosen = $${}^6{C_4}$$ ways

And from 4 different dictionaries 1 can be chosen = $${}^3{C_1}$$ ways

$$\therefore$$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen = $${}^6{C_4} \times {}^3{C_1}$$ ways

Let 4 novels are N1, N2, N3, N4 and 1 dictionary is D1.

Dictionary should be in the middle. So the arrangement will be like this

_ _ D1 _ _

On those 4 blank places 4 novels N1, N2, N3, N4 can be placed. And 4 novels can be arrange $$4!$$ ways.

$$\therefore$$ Total no of ways = $${}^6{C_4} \times {}^3{C_1}$$$$ \times 4!$$ = 1080

Questions Asked from Permutations and Combinations

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