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1

### AIEEE 2012

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is:
A
880
B
629
C
630
D
879

## Explanation

For alike n objects, number of ways we can select zero or more objects = n + 1 and number of ways we can select at least one object = n

Given 10 identical white balls, 9 identical green balls and 7 identical black balls.

To find number of ways for selecting atleast one ball.

Number of ways to choose zero or more white balls = (10 + 1) [since, all white balls are mutually identical]

Number of ways to choose zero or more green balls = (9 + 1) [since, all green balls are mutually identical]

Number of ways to choose zero or more black balls = (7 + 1) [since, all black balls are mutually identical]

Hence, number of ways to choose zero or more balls of any colour = (10 + 1) (9 + 1) (7 + 1)

Also, number of ways to choose a total of zero balls = 1

Hence, the number, if ways to choose at least one ball (irrespective of any colour) = (10 + 1) (9 + 1) (7 + 1) $$-$$ 1 = 879

2

### AIEEE 2011

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.

3

### AIEEE 2011

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 $${}^{6}{C_3}$$.

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

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

$$= 120 - 20$$

$$= 100$$

4

### AIEEE 2010

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

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