Permutation and Combination Exercise

Treat B and T as a single letter. Then the remaining letters (5 + 1 = 6) can be arranged in 6! ways. Since, O is repeated twice, we have to divide by 2 and the B and T letters can be arranged in 2! ways.

Total no. of ways

Required number of ways

Required number

= 4! × 2! = 24× 2 = 48

The word SIGNATURE consists of nine letters comprising four vowels (A, E, I and U) and five consonants (G, N, R, T and S). When the four vowels are considered as one letter, we have six letters which can be arranged in ⁶P₆ ways ie 6! ways. Note that the four vowels can be arranged in 4! ways.

Hence required number of words = 6! × 4! = 720 × 24 = 17280

Starting with the letter A, and arranging the other four letters, there are 4! = 24 words.

These are the first 24 words.

Then starting with G, and arranging A, A, I, and N in different ways, there are

words.

Hence, total 36 words.

Next, the 37th word starts with I.

There are 12 words starting with I.

This accounts up to the 48th word.

The 49th word is NAAGI. The 50th word is NAAIG.

No. of words starting with A are 4 ! = 24

No. of words starting with H are 4 ! = 24

No. of words starting with L are 4 ! = 24

These account for 72 words

Next word is rahlu and the 74th word rahul.

Total number of arrangements of letters in the word GARDEN = 6 ! = 720.

There are two vowels A and E, in half of the arrangements A precedes E and other half A follows E.

So, vowels in alphabetical order in

Here, 5 men out of 8 men and 6 women out of 10 women can be chosen in

⁸C₅ × ¹⁰C₆ ways

i.e., 11760 ways.

Since each bulb has two choices, either switched on or off.

Therefore required number = 2¹⁰ – 1 = 1023.

Here we have to divide 52 cards into 4 sets, three of them having 17 cards each and the fourth one having just one card.
First we divide 52 cards into two groups of 1 card and 51 cards.
This can be done in ways.
Now every group of 51 cards can be divided into 3 groups of 17 each in
.
Hence the required number of ways
=

Required number of possible outcomes

= Total number of possible outcomes – Number of possible outcomes in which 5 does

not appear on any dice. (hence 5 possibilities in each throw)

= 6³ – 5³ = 216 – 125 = 91

Ten candidates can be ranked in 10! ways.
In half of these ways A₁ is above A₂ and in another half A₂ is above A₁.
So, required number of ways is .

Let the two boxes be B₁ and B₂.

There are two choices for each of the n objects.

So, the total number of ways is

.

The number of ways of selecting 3 persons from 12 people under the given condition :

Number of ways of arranging 3 people among 9 people seated in a row, so that no two of them are consecutive

= Number of ways of choosing 3 places out of the 10 [8 in between and 2 extremes]

=

Total number of words that can be formed = 10⁵.

Number of words in which no letter is repeated = ¹⁰P₅.

So, number of words in which at least one letter is repeated

= 10⁵ – ¹⁰P₅ = 69760.

If a number is divisible by 3, the sum of the digits in it must be a multiple of 3.

The sum of the given six numerals is 0 + 1 + 2 + 3 + 4 + 5 = 15.

So to make a five digit number divisible by 3 we can either exclude 0 or 3.

If 0 is left out, then 5! = 120 number of ways are possible.

If 3 is left out, then the number of ways of making a five digit numbers is 4 × 4! = 96, because 0 cannot be placed in the first place from left, as it will give a number of four digits.

Hence, total number = 120 + 96 = 216.

Total number of ways

= 6 × 6 × ..... to n times = 6n.

Total number of ways to show only

even number = 3 × 3 × ...... to n times = 3n.

∴ required number of ways = 6n – 3n.

At least one black ball can be drawn in the following ways:

(i) one black and two other colour balls

(ii) two black and one other colour balls, and

(iii) all the three black balls

Therefore the required number of ways is

Required no. of the ways

= ⁶C₃ × ⁴C₂ = 20 × 6 = 120

No. of ways in which 6 men can be arranged at a round table = (6 - 1)!

Now women can be arranged in 6! ways.

Total Number of ways = 6! × 5!

Required number of ways

= coefficient of x² in (x + x² + ... x⁶)³

[Since each box can receive minimum 1 and maximum 6 balls]

= coefficient of x⁸ in x²(x + x + x² + ... + x⁵)³

= coefficient of x⁵ in

= coefficient of x⁵ in (1 - x)-³

= coefficient of

= ⁷C₅ = 21

Principal can be appointed in 36 ways.

Vice principal can be appointed in the remaining 35 ways.

Total number of ways = 36 × 35 = 1260