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1

### JEE Main 2015 (Offline)

The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is:
A
120
B
72
C
216
D
192

## Explanation

For a four digit number the first place can be filled in 3 ways with 6 or 7 or 8 and the remaining four places in 4! ways i.e., 3 $$\times$$ 4! = 72.

For a five digit number it can be arranged in 5! ways,

$$\therefore$$ total number of integers = (72 + 120) = 192.

2

### JEE Main 2013 (Offline)

Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A $$\times$$ B having 3 or more elements is -
A
219
B
211
C
256
D
220

## Explanation

A $$\times$$ B will have 2 $$\times$$ 4 = 8 elements.

The number of subsets having atleast 3 elements

= 8C3 + 8C4 + 8C5 + 8C6 + 8C7 + 8C8

= 28 – (8C0 + 8C1 + 8C2) = 256 – 1 – 8 – 28 = 219
3

### JEE Main 2013 (Offline)

Let $${T_n}$$ be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If $${T_{n + 1}} - {T_n}$$ = 10, then the value of n is :
A
7
B
5
C
10
D
8

## Explanation

Number of possible triangle using n vertices = nC3

$$\therefore$$ Tn = nC3

then Tn + 1 = n + 1C3

Given, $${T_{n + 1}} - {T_n}$$ = 10

$$\Rightarrow$$ n + 1C3 - nC3 = 10

$$\Rightarrow$$ $${{\left( {n + 1} \right)n\left( {n - 1} \right)} \over 6} - {{n\left( {n - 1} \right)\left( {n - 2} \right)} \over 6}$$ = 10

$$\Rightarrow$$ 3n(n - 1) = 60

$$\Rightarrow$$ n(n - 1) = 20

$$\Rightarrow$$ n2 - n - 20 = 0

$$\Rightarrow$$ (n - 5)(n + 4) = 0

$$\therefore$$ n = 5
4

### 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

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