# 66. Permutations and Combinations Solutions Part 3#

The word “Independence” has \(12\) letters with \(3\) ns, \(2\) ds, \(4\) es and one each of i, p and c.

Thus, total no. of permutations possible \(= \frac{12!}{3!2!4!}\)

The word “Principal” has \(9\) letters with:math:2 ps, \(2\) is, and one each of r, n, c, a and l.

Taking all three vowels together as one total no. of permutation is

\(\frac{7!}{2!}\) but the vowels themselves can be arranged in \(\frac{3!}{2!}\) ways.

\(\therefore\) desired answer \(= \frac{7!3!}{2!2!}\)

Mathematics has \(11\) letters with m coming twice, a coming twice, t coming twice, h, e, t, i, c, s each coming once.

Therefore, total no. of words that can be formed using these \(= \frac{11!}{2!2!}\)

Treating vowels and consonants together no. of words that can be formed \(= 2!. \frac{8!3!}{2!2!2!}\)

That is because \(8\) consonants can be arranged in \(\frac{8!}{2!2!}\) ways.

\(3\) vowels can be arranged in \(\frac{3!}{2!}\) ways.

Finally, the consonants and vowels can be arranged in \(2!\) ways when taken together.

The word “Director” has \(8\) letters with only \(r\) repeating for \(2\) times. Taking three vowels as one we can arrange them in \(\frac{6!}{2!}\) ways but three vowels can be arranged in \(3!\) ways among themselves.

Thus, desired answer \(= \frac{3!6!}{2!}\)

The word “Plantain” has \(8\) letters with a and n repeating \(2\) times each. Treating all vowels as one word we have total no. of arrangements as \(\frac{6!}{2!}\) but \(3\) vowels can be arranged in \(\frac{3!}{2!}\) ways among themselves.

Thus, desired answer \(= \frac{3!6!}{2!2!}\)

The word “Intermediate” has \(12\) letters with i coming twice, t coming twice, e coming thrice, n, r, m, d, a coming once.

So we have \(6\) consonants and \(6\) vowels.

Sinec the relative order does not change consonants will occupy consonants place and vowels will occupy vowels place.

Therefore, total no. of desired words \(= \frac{6!6!}{2!2!3!}\)

The word “Parallel” has \(8\) letters with l coming thrice, a coming twice, p, r, e each coming once.

Thus, total no. of arrangements \(= \frac{8!}{3!2!}\)

Treating all l as one; no. of numbers when all l come together \(= \frac{6!}{2!3!}\)

\(\therefore\) Desired answer \(= \frac{8!}{3!2!} - \frac{6!}{2!31}\)

Let us solve for each case one by one.

In this case position of D is fixed. So, remaining \(4\) letters can be arranged in \(4!\) ways as there is no repetition.

Same as previous case \(4!\) is the answer.

Same as previous case \(4!\) is the answer.

In this case position of two letters is fixed so answer is \(3!\)

The word “violent” has \(7\) letters. If vowels are to occupy all odd places then \(4\) consonants can be arranged in \(4!\) ways and vowels can be arranged in \(3!\) ways.

Thus, desired answer \(= 4!3!\)

The word \(Saloon\) has \(6\) letters with o repeating twice.

Since, vowels and consonants occupy alternate places total no. of such words \(= 2\frac{3!3!}{2!}\) as \(3\) consonants can be arranged in \(3!\) ways for \(3 places\) and vowels can be arranged in \(\frac{3!}{2!}\) ways. But then odd positions and even positions can be swapped doubling the no. of words formed.

The word \(Article\) has \(7\) letters with no repetitions.

Following as 109 total no. of such words \(= 4!3!\)

Since the number has to be greater than \(4\) million the most significant place can be occupied by either \(4\) or \(5\)

**Case I:**When most significant digit is \(4\)Rest of the digits can arrnaged in \(\frac{6!}{2!2!}\) ways as both \(3\) and \(5\) repeat twice.

**Case II:**When most significant digit is \(5\)Total no. of numbers \(= \frac{6!}{2!}\)

Therefore, desired answer \(= \frac{6!}{2!2!} + \frac{6!}{2!}\)

Total no. of numbers \(= \frac{7!}{3!2!}\) as \(2\) repeats thrice and \(3\) repeats twice.

For the no. of be odd unit’s place must be occupied by one of \(1, 3, 5\).

**Case I:**When unit’s place is occupied by \(1\) or \(5\)Total no. of numbers \(= 2 \frac{6!}{3!2!}\)

**Case II:**When unit’s place is occupied by \(3\)Total no. of numbers \(= \frac{6!}{3!}\)

Now total no. of odd numbers is sum of these two cases.

There are four odd places with \(4\) odd digits with both \(1\) and \(3\) reapeting twice.

Also, there are three even places with \(2\) repeating twice and \(4\) coming only once.

Thus, total no. of numbers \(= \frac{4!}{2!2!}\frac{3!}{2!}\)

**Case I:**When \(1\) occurs at most significant place and the number is of \(5\) digits.Remaining \(4\) positions can be filled in \(^5P_4\) ways.

**Case II:**When \(1\) occurs at most significant place and the number is of \(6\) digits.Remaining \(5\) positions can be filled in \(^5P_5\) ways.

**Case III:**When one of \(2, 3, 4, 5\) occurs at most significant place and the number is of \(5\) digits.Remaining \(4\) positions can be filled in \(\frac{^5P_4}{2!}\) ways.

Thus, total no. of such numbers \(= 4.\frac{^5P_4}{2!}\)

**Case IV:**When one of \(2, 3, 4, 5\) occurs at most significant place and the number is of \(6\) digits.Remaining \(5\) positions can be filled in \(\frac{5!}{2!}\) ways.

Thus, total no. of such numbers \(= 4.\frac{5!}{2!}\)

Desired answer is sum of these.

Most significant place can be filled using one of \(1, 2, 3, 4, 5\) i.e. in \(5\) ways. Rest of the places can be filled in \(6\) ways.

Thus, total no. of four digit numbers \(= 5\times 6^3\)

Total no. of four digit numbers without any digit repeating \(= 5\times 5\times 4\times 3\)

The difference of these two is the no. of numbers where at least one digit repeats.

Total no. of signals \(= \frac{9!}{5!2!2!}\)

Total no. of signals \(= ^6P_1 +~^6P_2 +~^6P_3 +~^6P_4 +~^6P_5 +~^6P_6\)

Total no. of arrangements \(= 5!\). In half of these \(e\) will come before \(i\).

\(4!\)

Let us solve each case:

\(9!\) because there is no restriction from the formula for circular permutation.

Let us first fix position of boys. They can be seated in \(4!\) ways from the formula of circular permutation.

This will make \(5\) empty seats between boys and girls can be seated in \(^5P_5\) ways.

Therefore, total no. \(= 4!5!\)

Treating all girls as a single entity total no. of seating arrangements \(= 6!\) from the formula of circular permutation as there will be \(7\) entities to be seated.

But the \(4\) girls can be seated in \(4!\) ways among themselves.

Thus, total no. of numbers \(= 6!4!\)

If the line starts with a girl then total no. of ways \(= 5!5!\)

If the line starts with a boy then total no. of ways \(= 5!5!\)

Thus, total no. of ways \(= 2.5!5!\)

**Second Part:**First let us make the boys take up places. So \(5\) boys can be arranged in \(4!\) ways. Now \(5\) girls can be seated in \(5\) empty seats in \(5!\) ways.Thus, total no. of seating arrangements \(= 4!5!\)

Following as last problem, total no. of ways \(= 5!\times~^6P_5\)

Since clockwise or anti-clockwise arrangement does not matter in a necklace total no. of ways \(= \frac{49!}{2}\)

Treating two particular delegated and host as one person we have \(19\) persons.

Total no. of ways of seating \(19\) persons around a round table \(= 18!\)

But the delegated and host can be arranged in \(2!\) ways as two delegated sit on either side of host.

Thus, total no. of desired arrangements \(= 18!2!\)

Four gentlemen can be seated in \(3!\) ways. This will make \(4\) empty seats between gentlemen so that no two ladies sit together.

No. of ways of making ladies sit in these \(4\) positions \(= 4!\)

Thus, total no. of arrangements \(= 3!4!\)

Following as previous problem; total no. of arrangements \(= 6!~^7P_6\)

We know that if \(^nC_x =~^nC_y\) then either \(x = y\) or \(x + y = n\)

\(\therefore\) Either \(3r = r+3\Rightarrow r = \frac{3}{2}\) which is not possible as \(r\) is an integer.

\(\therefore 3r + r + 3 = 15 \Rightarrow r = 3\)

\(^nC_6:^{n - 3}C_3 = 33:4\)

\(\therefore \frac{n!}{6!(n-6)!}\times\frac{3!(n - 3 - 3)}{(n-3)!} = \frac{33}{4}\)

\(\therefore \frac{n!}{(n - 3!)}.\frac{3!}{6!} = \frac{33}{4}\)

\(n(n - 1)(n - 2) = 6.5.33 = 11.10.9\)

\(n = 11\)

Given expression is \(^{47}C_4 + (^{51}C_3 + ^{50}C_3 + ^{49}C_3 + ^{48}C_3 + ^{47}C_3)\)

\(= ^{48C_4} + (^{51}C_3 + ^{50}C_3 + ^{49}C_3 + ^{48}C_3) [\because ^{47C_4 + ^{47}C_3} = ^{48}C_4]\)

Following similarly, \(= ^{52}C_4\)

Let \(p\) be the product of \(r\) consecutive integers beginning with \(n\)

Then \(p = n(n + 1)(n + 2) \ldots (n + r - 1)\)

\(\frac{p}{r!} = \frac{n(n + 1)(n + 2) \ldots (n + r - 1)}{r!}\)

\(= ^{n + r - 1}C_r\)

\(= ` number of combinations of :math:`n + r - 1\) things taken \(r\) at a time = an integer

Therefore, product is divisible by \(r!\)

Total no. of triangles formed \(= ^mC_3\)

Total no. of ways of choosing \(3\) children out of \(8 = ^8C_3 = 56\)

Therefore, the man will have to go to zoo for \(56\) times.

**Second Part:**Choosing \(3\) children out of \(8\) when a particular child is always chosen \(= ^1C_1.^7C_2 = 21\)Choosing a pair of students out of \(n = ^nC_2\)

Total no. of cards sent by one pair \(= 2\)

Therefore, total no. of cards send \(= 2.^nC_2 = 600\)

\(n^2 - n - 600 = 0\)

\(n = -24, 25\)

But \(n\) cannot be negative. Thus, \(n = 25\)

Whenever two vertices are joined either a diagonal or a side is formed.

Total no. of diagonals and sides \(= ^mC_2\)

But out of these \(m\) will be sides.

Therefore, no. of diagonals \(= ^mC_2 - m = \frac{m(m - 1)}{2} - m\)

\(= \frac{m(m - 3)}{2}\)

Total no. of persons \(= 10\)

Choosing \(5\) persons out of \(10 = ^{10}C_5\)

Choosing \(5\) persons out of \(6\) men \(= ^6C_5\)

Thus, no. of committees where at least one women is selected \(= ^10C_5 - ^6C_5\)

Let us solve these one by one:

Total no. of selections of \(3\) points out of \(10 = ^{10}C_3\)

Total no. of selection of \(4\) points out of \(4\) collinear points \(= ^4C_3\)

Thus, no. of triangles formed \(= ^{10}C_3 - ^4C_3\)

Total no. of selections of \(2\) points out of \(10\) points \(= ^{10}C_2 = 45\)

Number of selections of two points when only one straight line is formed \(= ^4C_2 = 6\)

i.e. the two points are selected from \(4\) collinear points making the straight line only one.

Therefore, total no. of straight lines \(= 45 - 6 + 1 = 40\)

Total no. of selections of \(4\) points out of \(10 = ^{10}C_4 = 210\)

Number of selections when no quadrilateral is formed \(= ^4C_3.^6C_1 + ^4C_4.6C_0 = 25\)

Thus, total no. of quadrilaterals formed \(= 210 - 25 = 185\)

**Case I:**When fruits of same type are different.Number of selections \(= (^4C_0 + ^4C_1 + \ldots + ^4C_4)\) \((^5C_0 + ^5C_1 + \ldots + ^5C_5)(^6C_0 + ^6C_1 + \ldots + ^6C_6) - 1\)

\(= 2^4.2^5.2^6 - 1 = 2^15 - 1\)

**Case I:**When fruits of same type are same.Zero or more oranges can be selected in \(5\) ways.

Zero or more apples can be selected in \(6\) ways.

Zero or more mangoes can be selected in \(7\) ways.

Thus, total no. of ways \(= 5.6.7 = 210\)

But in one of these ways no fruit is selected, making answer \(209\).

Following case I of previous problem we have total no. of ways \(= (2^5 - 1).(2^4 - 1).2^3 - 1 = 3720\)

Factors of \(21,600\) are \(2, 2, 2, 2, 2, 3, 3, 3, 5, 5\)

Thus, total no. of divisors \(= 6.4.3 - 1 = 71\)

Total no. of divisors including \(1 = 72\)

No. of ways \(= ^5C_1 + ^5C_2 + ^5C_3 + ^5C_4 + ^5C_5 = 2^5 - 1 = 31\)

Required number \(= ^{12}C_4.^8C_4.^4C_4\)

Total no. of letters in the word “EXAMINATION” \(= 11\) out of which A, N, I occur twice. Different letters are E, X, A, M, I, N, T, O which are eight in count.

**Case I:**When all are different.Total no. of words formed \(= ^8C_4.4!\)

**Case II**When two are identical and two are different.There are three possibilities of identical letters with A, N, I each occurring twice.

Total no. of words formed \(= 3.^7C_2.\frac{4!}{2!} = 756\)

**Case III:**When letters are pair of identical letters.Total no. of words formed \(= 3.\frac{4!}{2!2!} = 18\)

\(\because 30 \neq 4; n = 30 + 4 = 34\)

\(\because 12 \neq 8; n = 12 + 8 = 20\)

\(^{20}C_{17} = \frac{20.19.18}{3.2} = 1140\)

\(^{22}C_{20} = \frac{22.21}{2} = 231\)

\(\because r \neq r + 2; r + r + 2 = 18 \Rightarrow r = 8\)

\(^8C_6 = 28\)

\(n(n - 1)(n - 2)(n - 3) = 15.4.3.2\)

\(n(n - 1)(n - 2)(n - 3) = 6.5.4.3 \Rightarrow n = 6\)

\(\frac{15!}{r!(15-r)!}.\frac{(r-1)!(16 - r)!}{15!} = 11:5\)

\(\Rightarrow r = 5\)

\(r = 5\)