Permutation and combination-Exercise 1.2

1.Please see the question from book.

Soln:

Here, n = 5 , r = 3.

The permutation of 5 objects taken 3 at a time = P(n,r)

= P(5,3) = 5!(5−3)!$\frac{5!}{\left(5-3\right)!}$ = 5!2!$\frac{5!}{2!}$ = 5 * 4 * 3 = 60 ways.

2.Please see the question from book.

Soln:

Here, n = 10, r = 3

3 persons in 10 vacant seats can be arranged in P(10,3) different ways

= 10!(10−3)!$\frac{10!}{\left(10-3\right)!}$ = 10!7!$\frac{10!}{7!}$ = 10 * 9 * 8 = 720 ways.

3.Please see the question from book.

Soln:

Here, given number of objects (n) = 6

Number in each arrangement (r) = 4.

Then, P(6,4) = 6!(6−4)!$\frac{6!}{\left(6-4\right)!}$ = 6∗5∗4∗3∗2∗12∗1$\frac{6\ast 5\ast 4\ast 3\ast 2\ast 1}{2\ast 1}$ = 360

So, there are 360 plates of vehicles can be made.

4.Please see the question from book.

a.

Soln:

If the boys and girls may sit anywhere, then there are 7 persons and 7 seats. 7 persons in 7 seats can be arranged in P(7,7) ways.

= 7!(7−7)!$\frac{7!}{\left(7-7\right)!}$ = 7!0!$\frac{7!}{0!}$ = 7∗6∗5∗4∗3∗2∗11$\frac{7\ast 6\ast 5\ast 4\ast 3\ast 2\ast 1}{1}$ = 5,040 ways.

b.

Soln:

If the boys and girls must sit alternately, there are 4 seats for boys and 3 for girls.

Here, for boys n = 4, r = 4

4 boys in 4 seats can be arranged in P(4,4) ways

= 4!(4−4)!$\frac{4!}{\left(4-4\right)!}$ = 4∗3∗2∗11$\frac{4\ast 3\ast 2\ast 1}{1}$ = 24 ways.

Again. For girls n = 3, r = 3

3 girls in 3 seats can be arranged in P(n,r) i.e. P(3,3) ways.

= 3!0!$\frac{3!}{0!}$ = 3∗2∗11$\frac{3\ast 2\ast 1}{1}$ = 6ways.

So, total no. of arrangement = 24 * 6 = 144 ways.

c.

Suppose 3 girls = 1 object, then total number of student (n) = 4 + 1= 5.

Then the permutation of 5 objects taken 5 at a time.

= P(5,5) = 5!(5−5)!$\frac{5!}{\left(5-5\right)!}$ = 5∗4∗3∗2∗11$\frac{5\ast 4\ast 3\ast 2\ast 1}{1}$ = 120.

We know, 3 girls can be arranged themselves in P(3,3) different ways,

i.e. P(3,3) = 3!(3−3)!$\frac{3!}{\left(3-3\right)!}$ = 3 * 2 * 1 = 6 different ways.

Therefore, required of arrangements = 120 * 6 = 720.

5.Please see the question from book.

Soln:

Two people have to insist on sitting next to each other, so two people are considered as one.

So, n = 7, r = 7

7 people in 7 seats can be arranged in P(7,7) ways

= 7!0!$\frac{7!}{0!}$ = 7∗6∗5∗4∗3∗2∗11$\frac{7\ast 6\ast 5\ast 4\ast 3\ast 2\ast 1}{1}$ = 5,040 ways.

The two people can interchange their position in 2 ways.

So, total no. of arrangements = 2 * 5040 = 10,080 ways.

6.Please see the question from book.

Soln:

Total number of arrangements of 6 objects taken 6 at a time.

= P(6,6) = 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720.

a.

Suppose two particular books = 1object, which are always come together then number of objects = 5.

Now, the number of arrangements of 5 objects taken 5 at a time = 5! = 5 * 4 * 3 * 2 * 1 = 120.

2 books can be arranged themselves in 2! = 2 * 1 = 2 different ways.

Then the required number of arrangements = 120 * 2 = 240

b.

The required number of arrangements of books when two particular books are not come together = 720 – 240 = 480.

7.Please see the question from book.

Soln:

There are 4 + 5 + 3 = 12 beads in total of which 4 are red, 5 white and 3 blue, n = 12, p = 4, q = 5, r = 3.

So, total no. of arrangement = n!p!.q!.r!$\frac{\mathrm{n}!}{\mathrm{p}!.\mathrm{q}!.\mathrm{r}!}$ = 12!4!.5!.3!$\frac{12!}{4!.5!.3!}$ ways.

8.Please see the question from book.

Soln:

a.

There are 7 letters in word “ELEMENT”. Also, there are 3 E’s and rests are different.

Then the total number of arrangements = 7!3!$\frac{7!}{3!}$ = 7∗6∗5∗4∗3!3!$\frac{7\ast 6\ast 5\ast 4\ast 3!}{3!}$ = 840.

b.

There are 8 letters in the word “NOTATION”. Also there are 2 N’s, 2 T’s, 2 O’s and rests are different.

So, the total number of arrangements = 8!2!.2!.2!$\frac{8!}{2!.2!.2!}$ = 8∗7∗6∗5∗4∗3∗2∗12∗1∗2∗1∗2∗1$\frac{8\ast 7\ast 6\ast 5\ast 4\ast 3\ast 2\ast 1}{2\ast 1\ast 2\ast 1\ast 2\ast 1}$ = 6720.

c.

There are 11 letters in the word “MATHEMATICS”. There are 2 M’s, 2 A’s and 2 T’s and rest are different.

d.

There are 11 letters in word “MISSISSIPPI. There are 4 I’s, 4 S’s. 2 P’s and rest are different.

Then the required number of arrangement = 11!4!.2!.2!$\frac{11!}{4!.2!.2!}$

9.Please see the question from book.

Soln:

There are 6 objects out of which there are 2 2’s and 3 3’s. Then the number of permutation = 6!2!.3!$\frac{6!}{2!.3!}$ = 6∗5∗4∗3∗2∗12∗1∗3∗2∗1$\frac{6\ast 5\ast 4\ast 3\ast 2\ast 1}{2\ast 1\ast 3\ast 2\ast 1}$ = 60.

But for 6 digit number, 0 can not occur in the 1st place. So there are only 5 digits left.

The number of 6 digit number with 0 in the 1st place with repetition of 2.2’s and 3 3’s is 5!2!.3!$\frac{5!}{2!.3!}$ = 5∗4∗3∗2∗12∗1∗3∗2∗1$\frac{5\ast 4\ast 3\ast 2\ast 1}{2\ast 1\ast 3\ast 2\ast 1}$ = 10.

So, the required number of 6 digit significant numbers = 60 – 10 = 50.

10.Please see the question from book.

Soln:

a.

Total number of objects = 4 + 4 = 8.

If they may sit anywhere, then it is the circular arrangements of 8 objects taken 8 at a time. So, the total number of permutation.

= (8 – 1)! = 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040.

b.

Art and Science students have to sit alternately in a round table. So, there are 4 seats for Art students and 4 for Science students.

4 art students at a round table can be arranged in 4 – 1! ways.

= 3! = 3 * 2 * 1 = 6 ways.

Again. 4 science students can be arranged in P(4,4) ways, i.e. 4! Ways = 4 * 3 * 2 * 1 = 24 ways.

So, total no. of arrangement = 6 * 24 = 144 ways.

11.Please see the question from book.

Soln:

Two people have to insist on sitting next to each other, so two people are considered as one. Then the 7 people can be arranged around a table in 7 – 1! = 6! ways. i.e. 6!

= 6 * 5 * 4 * 3 * 2 * 1 = 720 ways.

The two people can interchange their position in 2 ways.

So, total no. of arrangement = 2 * 720 = 1,440 ways.

12.Please see the question from book.

Soln:

7 beats in bracelet can be arranged in 7 – 1! ways i.e. 6 ! ways.

= 6 * 5 * 4 * 3 * 2 * 1 = 720 ways.

Since, the distinction between clockwise and anti clockwise is not made, so the required arrangements

= 12$\frac{1}{2}$.720 = 360 ways.

13.Please see the question from book.

Soln:

a.

There is n = 6 and r = 4.

Then the number of permutation = nr = 64 = 1296 ways.

b.

There are 10 objects (numbers) to form 3 digits numbers. To form even numbers there are 5 ways to fill up the unit place, there are 9 ways to fill up the 100th place and there are 10 ways to fill up the 10th place of the 3 digit numbers (when repetition is allowed.)

From basic principle of counting, required number of permutation = 9 * 10 * 450  [0 cannot be put in the hundred place.]

c.

If each students receive any number of prizes (i.e. repetition is allowed).

Then the 3 prizes can be distributed among 4 students.

= 4 * 4 * 4 different ways = 64 different ways.

14.Please see the question from book.

Soln:
The letters of the word “Monday” can be arranged in P(6,6) ways

= 6!0!$\frac{6!}{0!}$ = 6∗5∗4∗3∗2∗11$\frac{6\ast 5\ast 4\ast 3\ast 2\ast 1}{1}$

= 720 ways.
 

To find the arrangements of the letters of the word ‘Monday” always beginning with M, we fix M. The remaining 5 letters can be arranged in P(5,5) ways  5!0!$\frac{5!}{0!}$ = 5∗4∗3∗2∗11$\frac{5\ast 4\ast 3\ast 2\ast 1}{1}$ = 120 ways.

So, no of arrangements of the letter of the word “Monday” not beginning with M = 720 – 120 = 600.

To find the arrangements beginning with M and ending with y, we fix M and y. The remaining 4 letters can be arranged in P(4,4) ways = 4!0!$\frac{4!}{0!}$ = 4∗3∗2∗11$\frac{4\ast 3\ast 2\ast 1}{1}$ = 24 ways.

So, no of arrangements of the letters of the word “Monday” beginning with M and not ending with y = 120 – 24 = 96.

15.Please see the question from book.

Soln:

a.

There are 7 letters in the word “COLLEGE”.

There are 2 L’s, 2 E’s and rest are different.

Let 2 E = 1object(when they come together).

Then number of objects  = 6.

So, the required number of permutation = 6!2!$\frac{6!}{2!}$ = 360.

b.

There are 7 letters in word “ARRANGE”. There are 2 A’s, 2 R’s and rest are different.

Therefore, number of permutation = 7!2!.2!$\frac{7!}{2!.2!}$ = 1260.

To find the arrangements always coming 2 R’s together, we consider 2 R’s as one. Then the arrangements with 2 R’s always together = 6!2!$\frac{6!}{2!}$ = 360

So, the number of arrangements of no two R’ come together = 120 – 360 = 900.

16.Please see the question from book.

a.

Soln:

There are 3 vowels and 5 consonants.

Suppose 3 vowels = 1 object, then the number of object = 5 + 1 = 6.

Permutation of 6 objects taken 6 at a time = P(6,6)

= 6 * 5 * 4 * 3 * 2 * 1 = 720.

3 vowels can arrange themselves in 3! Different ways

= 3 * 2 * 1 = 6 different ways.

So, total number of arrangements = 720 * 6 = 4320.

b.

There are 3 vowels and 4 positions for them. So, 3 vowels can be arranged in P(4,3) = 4!(4−3)!$\frac{4!}{\left(4-3\right)!}$ = 4 * 3 * 2 * 1 = 24 different ways.

Also, the remaining 5 letters (i.e. C, M, P, T, R) can be arranged in P(5,5) = 5!(5−5)!$\frac{5!}{\left(5-5\right)!}$ = 5 * 4 * 3 * 2 * 1 = 120 different ways.

So, total number of arrangements = 24 * 120 = 2880.

c.

The three vowels can be arranged in 3! Different ways and the five consonants can be arranged in 5! different ways.

So, total number of arrangements = 3 ! * 5 ! = 3 * 2 * 1 * 5 * 4 * 3 * 2 * 1 = 720.