Permutations and Combinations

To find the number of ways to arrange 10 friends at dinner on two round tables, one with 4 people and the other with 6, follow these steps:

Divide into Groups:

You need to create two groups from 10 people: one group with 4 people and another with 6 people.

The number of ways to form these groups is given by the combination formula:

$ \frac{10!}{4! \times 6!} $

Arrange on Round Tables:

Arranging 4 people on a round table can be done in $(4-1)!$ ways (since arrangements around a circle account for rotations being identical):

$ 3! $

Arranging 6 people on a round table can be done in $(6-1)!$ ways:

$ 5! $

Total Arrangements:

Multiply the number of ways to form the groups by the number of ways to arrange each group on their respective tables:

$ \frac{10!}{4! \times 6!} \times 3! \times 5! $

Simplifying, this expression calculates to:

$ \frac{10!}{24} $

Hence, the total number of ways to arrange the guests is $\frac{10!}{24}$.

To solve the problem, we need to arrange the digits of the number 223355888 so that the odd digits occupy even positions. We have:

Odd digits: $3, 3, 5, 5$

Even digits: $2, 2, 8, 8, 8$

The number positions can be represented as:

$ \bar{O} \bar{E} \bar{O} \bar{E} \bar{O} \bar{E} \bar{O} \bar{E} \bar{O} $

Where $E$ indicates even positions and $O$ indicates odd positions.

Steps:

Arranging Odd Digits on Even Positions:

There are 4 even positions available for the odd digits $ (3, 3, 5, 5) $. The number of ways to arrange these with repetitions is given by:

$ \frac{4!}{2!2!} $

Arranging Even Digits on Odd Positions:

There are 5 odd positions available for the even digits $ (2, 2, 8, 8, 8) $. The number of ways to arrange these with repetitions is given by:

$ \frac{5!}{2!3!} $

Calculating Total Arrangements:

By multiplying the number of ways to arrange the odd and even digits, we get the total number of 9-digit numbers that can be formed:

$ \frac{4!}{2!2!} \times \frac{5!}{2!3!} = 60 $

Thus, there are $60$ different ways to arrange the digits to meet the given conditions.

To make 6-digit numbers from given digits $1,3,5, 7$ and 9 , we must repeat only one digit and we can done the same in ${ }^5 C_1$ ways.

Now, the arrangement of these 6 -digits in which two are identical is 6!/ 2.

So, required numbers of different 6-digit numbers

$={ }^5 C_1 \frac{6!}{2!}=5 \times \frac{720}{2}=1800$

Given word is LETTER, having vowels E, E and consonants L, T, T, R. Now, the number of ways to arrange the consonants are $\frac{4!}{2!}=12$

Now, we have five place to put vowel E, E.

So, number of ways to arrange vowels is ${ }^5 C_2=10$

So, number of required words $=12 \times 10=120$

Other than 2, remaining five places can be filled by 1 and 3 for each place. The number of ways for five places is 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 2 = 2⁵. For 2, selecting 2 places out of 7 is ⁷C₂. Hence, the required number of ways is ⁷C₂ $\times$ 2⁵.

The word HAVANA has 6 letters. However, the letters 'A' appear 3 times. We'll first consider all permutations, and then subtract the ones where 'V' and 'N' appear together.

1. Total permutations :

Since we have 6 total letters and 3 of them are the same (A), we use the formula for permutations of multiset to get the total number of arrangements, which is 6! / 3! = 720 / 6 = 120.

2. Permutations with 'V' and 'N' together :

Let's consider 'V' and 'N' as a single entity. So, we have 5 entities - {V,N}, 'A', 'A', 'A', 'H'.

Since 'A' appears 3 times, the total permutations would be 5! / 3! = 120 / 6 = 20.

However, 'V' and 'N' can be arranged among themselves in 2! = 2 ways.

So, total arrangements where 'V' and 'N' are together is 20 $ \times $ 2 = 40.

The total permutations where 'V' and 'N' do not appear together is the total permutations minus the ones where they are together: 120 - 40 = 80.

Two circles intersect maximum at two distinct points.

Now, two circles can be selected in ¹⁰C₂ ways.

The total number of points of intersection are ¹⁰C₂ $\times$ 2

$ = {{10 \times 9} \over {1 \times 2}} \times 2 = 90$

Possible even digits are 2, 4, 6, 8, 0.

Case I Number has digits 4, 6, 8, 0.

(Here, sum of digits 18, divisible by 3)

Number of arrangements = 3 $\times$ 3!

[1st place can be filled using 4, 6, 8]

= 3 $\times$ 6 = 18

Case II Number has digits 2, 4, 6, 0

(Here, sum of digits is 12, divisible by 3)

1st place cannot be filled by 0.

Number of arrangements = 3 $\times$ 3! = 18

$\therefore$ Number of numbers = 18 + 18 = 36

Let's denote the three digits of the number as $a $, $b$, and $ c $, and thus the number we're looking at is $100a + 10b + c$. According to the property described, the middle digit $b $ is the arithmetic mean of $a $ and $c$, which means:

$ b = \frac{a+c}{2} $

In order for $b$ to be an integer, $a + c$ must be an even number since the sum of two numbers divided by 2 can only result in an integer if the sum is even.

Since $a$, $b$, and $c$ are digits, they can only take values from 0 to 9, inclusive. However, for $a$, it cannot be 0 because we are looking for a three-digit number, which means $a$ must be from 1 to 9.

Let's consider all possible even $a + c$ sums:

$a = 1$ allows $c = 1, 3, 5, 7, 9$ (since $1 + c$ must be even and $c$ can be from 0 to 9).

$a = 2$ allows $c = 0, 2, 4, 6, 8$ (for the same reason).

$a = 3$ allows $c = 1, 3, 5, 7, 9$.

...

$a = 9$ allows $c = 1, 3, 5, 7, 9$.

So for each odd $a$, there are 5 possible values for $c$, and for each even $a$, there are also 5 possible values for $c$, totaling 5 choices for $c$ for each value of $a$.

Since $a$ can be from 1 to 9, there are, in total:

$ 9 \times 5 = 45 $

possible three-digit numbers where the middle digit is the arithmetic mean of the first and last digits. Therefore, the correct answer is:

Option B: 45

With four 6-sided dics, there are possible combinations that can be rolled. Out of these, only 80 sum to 10 as the table shows,

$ 6 \times 6 \times 6 \times 6=6^4=1296 $