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QUESTION: 1

What is the value of ^{100}P_{2}?

Solution:

We know , ^{n}P_{r} = n ! /( n - r) !

So ^{100}P_{2 }= 100 ! / ( 100 - 2 ) !

= 100 ! / 98 !

= 100 x 99 x 98 ! / 98 !

= 9900

So Option D is correct answer.

QUESTION: 2

A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw?

Solution:

The bag contains 2 White, 3 Black and 4 Red balls.

So, total 9 balls are there in the bag; among them 3 are Black and 6 are non-Black balls.

Three balls can randomly be drawn in (9C3) = 84 ways.

1 Black and 2 non-Black balls can be drawn in (3C1)*(6C2) = 45 ways.

1 non-Black and 2 Black balls can be drawn in (6C1)*(3C2) = 18 ways.

3 Black balls can be drawn in (3C3) = 1 way.

So, three balls drawn in (45 + 18 + 1) = 64 ways will have at least one Black ball among the drawn ones.

QUESTION: 3

A coin is tossed 3 times. Find out the number of possible outcomes.

Solution:

For any multiple independent event, there are *n*^{m}

total possible outcomes, where n is the number of outcomes per event, and m is the number of such events.

So for a coin, discounting the unlikely event of landing on its side, there are two possible outcomes per event, heads or tails. And it is stated that there are 3 such events. So *n*^{m}=2^{3}=8

.

QUESTION: 4

In how many ways can the letters of the word 'LEADER' be arranged?

Solution:

The word 'LEADER' has 6 letters.

But in these 6 letters, 'E' occurs 2 times and rest of the letters are different.

Hence,number of ways to arrange these letters

QUESTION: 5

How many 3 digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated?

Solution:

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

∴ Required number of numbers = (1 x 5 x 4) = 20.

QUESTION: 6

How many words with or without meaning, can be formed by using all the letters of the word, 'DELHI' using each letter exactly once?

Solution:

Explanation :

The word 'DELHI' has 5 letters and all these letters are different.

Total words (with or without meaning) formed by using all these

5 letters using each letter exactly once

= Number of arrangements of 5 letters taken all at a time

= ^{5}P_{5} = 5! = 5 x 4 x 3 x 2 x 1 = 120

QUESTION: 7

In how many different ways can the letters of the word 'JUDGE' be arranged such that the vowels always come together?

Solution:

The given word contains 5 different letters.

Keeping the vowels UE together, we suppose them as 1 letter.

Then, we have to arrange the letters JDG (UE).

Now, we have to arrange in 4! = 24 ways.

The vowels (UE) can be arranged among themselves in 2 ways.

∴ Required number of ways = (24 × 2) = 48

QUESTION: 8

How many arrangements can be made out of the letters of the word 'ENGINEERING' ?

Solution:

The number of arrangements of the word ENGINEERING is 277200.

Solution:

ENGINEERING word has 3 times of 3, three times of N, 2 times of G and 2 times of I. Then, the total letter is 11.

So, the number of arrangements of the word ENGINEERING = 11!/[3! * 3! * 2! * 2!]

= 39916800/[6 * 6 * 2 * 2]

= 277200

Hence, the number of arrangements of the word ENGINEERING is 277200.

QUESTION: 9

How many words can be formed by using all letters of the word 'BIHAR'?

Solution:

Explanation :

The word 'BIHAR' has 5 letters and all these 5 letters are different.

Total words formed by using all these 5 letters = ^{5}P_{5} = 5!

= 5 x 4 x 3 x 2 x 1 = 120

QUESTION: 10

In how many different ways can the letters of the word 'DETAIL' be arranged such that the vowels must occupy only the odd positions?

Solution:

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

(1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.

Number of ways of arranging the vowels = 3*P*3

= 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = 3*P*3

= 3! = 6.

Total number of ways = (6 x 6) = 36.

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