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1

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
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)

MCQ (Single Correct Answer)
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)

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

Questions Asked from Permutations and Combinations

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