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Q:

The Number of times the digit 8 will be written when listing the integers from 1 to 1000 is :

A) 100 B) 200
C) 300 D) 400

Answer:   C) 300



Explanation:

 Since 8 does not occur in 1000, we have to count the number of times 8 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form xyz, where \inline 0\leq x, y,z\leq 9.

Let us first count the numbers in which 8 occurs exactly once.

Since 8 can occur atone place in \inline ^{3}\textrm{C}_{1} ways. There are \inline ^{3}\textrm{C}_{1}(9\times 9)=3\times 9^{2} such numbers.

Next, 8 can occur in exactly two places in \inline ^{3}\textrm{C}_{2}(9)=3\times 9 such numbers. Lastly, 8 can occur in all three digits in one number only.

Hence, the number of times 8 occur is 

\inline 1\times (3\times 9^{2})+2\times (3\times 9)+3\times 1=300

Q:

How many four digit even numbers can be formed using the digits {2, 7, 5, 3, 9, 1} ?

A) 59 B) 60
C) 61 D) 64
 
Answer & Explanation Answer: B) 60

Explanation:

The given digits are 1, 2, 3, 5, 7, 9 

A number is even when its units digit is even. Of the given digits, two is the only even digit.
Units place is filled with only '2' and the remaining three places can be filled in ⁵P₃ ways.
Number of even numbers = ⁵P₃ = 60.
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1 49
Q:

A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are picked at random, what is the probability that they are either blue or yellow ?

A) 3/17 B) 4/21
C) 2/21 D) 5/17
 
Answer & Explanation Answer: C) 2/21

Explanation:

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 x 2)/(15 x 14) = 1/35
Probability that both are yellow = ²C₂/¹⁵C₂ = (2 x 1)/(15 x 14) = 1/105
Probability that one blue and other is yellow = (³C₁ x ²C₁)/¹⁵C₂ = (2 x 3 x 2)/(15 x 14) = 2/35

Required probability = 1/35 + 1/105 + 2/35
= 3/35 + 1/105 = 1/35(3 + 1/3)
= 10/(3 x 35) = 2/21

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4 120
Q:

A group consists of 4 men, 6 women and 5 children. In how many ways can 2 men , 3 women and 1 child selected from the given group ?

A) 600 B) 610
C) 609 D) 599
 
Answer & Explanation Answer: A) 600

Explanation:

Two men, three women and one child can be selected in ⁴C₂ x ⁶C₃ x ⁵C₁ ways

=

= 600 ways.

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1 143
Q:

The number of sequences in which 4 players can sing a song, so that the youngest player may not be the last is ?

A) 2580 B) 3687
C) 4320 D) 5460
 
Answer & Explanation Answer: C) 4320

Explanation:

Let 'Y' be the youngest player.

The last song can be sung by any of the remaining 3 players. The first 3 players can sing the song in (3!) ways.

The required number of ways = 3(3!) = 4320.

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3 150
Q:

A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most  ?

A) 215 B) 268
C) 254 D) 216
 
Answer & Explanation Answer: A) 215

Explanation:

Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 x 6 x 6 = 216. Of these attempts, one of them is a successful attempt.

Maximum number of unsuccessful attempts = 216 - 1 = 215.

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