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Class 11 RD Sharma Solutions- Chapter 17 Combinations- Exercise 17.1 | Set 1
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Last Updated : 11 Feb, 2021
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Question 15.  There are three copies each of 4 different books. In how many ways can they be arranged in a shelf?

Solution:   

Total books = 3 x 4 = 12 

Ways to arrange books = 12!

Need to compensate for extra ways included due to identically in some books :

The three copies of each book are identical 

So, for each different book, they have been included 3! times 

So, for the different 4 books, they have been included 3! x 3! x 3! x 3! = 3!4 times

Hence, the ways to arrange the given books on the shelf = 12! / 3!4 

Question 16. How many different arrangements can be made by using all the letters of the word ‘MATHEMATICS’. How many of them begin with C? How many of them begin with T? 

Solution:   

Given: In ‘MATHEMATICS’ word

M appear = twice

T appear = twice

A appear = twice

Remaining letters = once 

So the total number of letters in the ‘MATHEMATICS’ word = 11 

Number of different arrangements = 11! / (2! x 2! x 2!) = 4989600

Beginning with C = fix C on position number 1 

 Arrange letters at the remaining positions 

= 10! / (2! x 2! x 2!) = 453600

Beginning with T = fix T on position number 1 

Now, the duplicate of T not left for remaining positions 

= 10! / (2! x 2!) = 907200

Question 17. A biologist studying the genetic code is interested to know the number of possible arrangements of 12 molecules in a chain. The chain contains 4 different molecules represented by the initials A (for Adenine), C(for Cytosine), G (for Guanine), and T (for Thymine), and 3 molecules of each kind. How many different such arrangements are possible?

Solution:   

Given: Total molecules = 12

Ways to arrange molecules = 12!

Need to compensate for extra ways included due to identically in some molecules:

4 kinds of molecules with 3 of each kind = 3! x 3! x 3! x 3! times 

Ways to arrange the given books on the shelf = 12!/ 3! x 3! x 3! x 3! = 369600   

Question 18. In how many ways can 4 red, 3 yellow, and 2 green discs be arranged in a row if the discs of the same color are indistinguishable?

Solution:   

Given: Total number of discs = 9

In which, 

Red color dice = 4

Yellow color dice = 3

Green color dice = 2

So, the total arrangements = 9! / (4! x 3! x 2!) = 1260

Question 19. How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4? 

Solution:   

All 7-digit numbers are greater than 1000000 with digits 1, 2, 0, 2, 4, 2, 4

For 1st digits = number of ways = 6 (except 0)

For 2nd digit = 6 ways (except digit on 1st)

For 3rd digit = 5 ways (except on 1st, 2nd)

And so on. 

Number of such numbers = 6 x 6 x 5 x 4 x 3 x 2 x 1 / (3! 2!)  (divided to remove duplicates in 2 and 4 digits) 

= 6 x 6! / (3! 2!) = 360

Question 20. In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?

Solution:   

Let, all S together can be assumed as 1 symbol = 10 letters left

In the given ASSASSINATION word

A appear = 3 times

N, I appear = 2 times

Number of such ways = 10! / (3! 2! 2!) = 151200

Question 21. Find the total number of permutations of the word ‘INSTITUTE’.

Solution:   

In the given word INSTITUTE

I appear = 2 times

T appear = 3 times

Hence, the number of permutations are = 9! / (2! x 3!) = 30240

Question 22. The letters of the word ‘SURITI’ are written in all possible orders and these words are written out as in a dictionary. Find the rank of the word ‘SURITI’.

Solution:   

 As we know that in dictionary the words in each stage are arranged in alphabetical order.

According to our problem we can consider that the word begin with I, R, S, T, and U

Number of words starting in I = 5!

Number of words starting in R = 5! / 2!

Number of words starting in SI = 4!

Number of words starting in SR = 4! / 2!

Number of words starting in ST = 4! / 2!

Number of words starting in SUI = 3!

Then the next word in dictionary is going to be = SURIIT

And next = SURITI

So, rank of SURITI = 5! + 5! / 2! + 4! + 4! / 2! + 4! / 2!+ 3! + 1 + 1 = 236

Question 23. If the letters of the word ‘LATE’ be permuted and the words so formed be arranged as in a dictionary, find the rank of the word LATE.

Solution:   

 As we know that in dictionary the words in each stage are arranged in alphabetical order.

According to our problem we can consider that the word begin with A, E, L, and T.

So, Words starting from A = 3!

Words starting from E = 3!

Now, word starting from L = 3!. But one of the word is LATE itself

So the first word is LAET and the next word is LATE

Hence, the rank of LATE is = 3! + 3! + 2 = 14    

Question 24. If the letters of the word ‘MOTHER’ are written in all possible orders and these words are written out as in a dictionary, find the rank of the word ‘MOTHER’.

Solution:   

 As we know that in dictionary the words in each stage are arranged in alphabetical order.

According to our problem we can consider that the word begin with E, H, M, O, T, and R.

So, words starting in E = 5!

Words starting in H = 5!

Words starting in ME = 4!

Words starting in MH = 4!

Words starting in MOE = 3!

Words starting in MOH = 3!

Words starting in MOR = 3!

Words starting in MOTE = 2!

MOTHER = next word  

Rank is = 5! + 5! + 4! + 4! + 3! + 3! + 3! + 2! + 1 = 309        

Question 25. If the permutations a, b, c, d, e taken all together be written down in alphabetical order as in dictionary and numbered, find the rank of the permutation debac.

Solution:   

 As we know that in dictionary the words in each stage are arranged in alphabetical order.

According to our problem we can consider that the word begin with a, b, c, d, and e.

So, the number of words starting in ‘a’ = 4!

Words starting in b = 4!

Words starting in c = 4!

Words starting in da = 3!

Words starting in db = 3!

Words starting in dc = 3!  

Words starting in dea = 2!

Next word = debac

Rank = 4! x 3 + 3! x 3 + 2! + 1 = 93    

Question 26. Find the total number of ways in which six ‘+’ and four ‘-‘ signs can be arranged in a line such that no two ‘-‘ signs occur together.

Solution:   

Total number of ‘-‘ sign = 4

Total number of ‘+’ sign = 6

The six ‘+’ signs arrange in a line = 1 way

Now, we have 7 places in which four different thing can be arranged but all the four ‘-‘ sign look identical 

so the ‘-‘ sign can be arranged = 7P4/4! = 35

Hence the number of ways = 1 x 35 = 35

Question 27. In how many ways can the letters of the word ‘INTERMEDIATE’ be arranged so that:

(i) The vowels always occupy even places?

(ii) The relative order of vowels and consonants do not alter?

Solution:   

(i) In the given word, total number of vowels are = 6 vowels

In this word there are total 6 possible even positions

Arrange the vowels on even positions = 6! / (2! x 3!) ways

Arrange consonants = 6! / 2! ways = 360

(ii) Vowels ordering = 6! / (2! x 3!)

Consonants ordering = 6! / 2!

Total permutations = 6! / (2! x 3!) x 6! / 2! = 21600  

Question 28. The letters of the word ‘ZENITH’ are written in all possible orders. How many words are possible if all those words are written out as in a dictionary? What is the rank of the word ‘ZENITH’?

Solution:   

 As we know that in dictionary the words in each stage are arranged in alphabetical order.

According to our problem we can consider that the word begin with E, H, I, N, T, and Z.

So, the total words possible are = 6!

Words starting in E = 5!

Words starting in H = 5!

Words starting in I = 5!

Words starting in N = 5!

Words starting in T = 5!

Words starting in ZEH = 3!

Words starting in ZEI = 3!

Words starting in ZENH = 2!

Words starting in ZENIH = 1!

ZENITH is next word

Rank = 5! x 5 + 3! x 2 + 2! + 1 + 1 = 616

Question 29. 18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?

Solution:

18 mice can be arranged themselves in 18P18 ways = 18!

According to the question, there are three groups, and they are equally large

So the 18 mice divided into three groups, and they can be arranged themselves inside the group 

Hence, the number of ways mice be placed into the three groups = 18!/6! x 6! x 6!

= 18!/(6!)3



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Class 11 RD Sharma Solutions- Chapter 17 Combinations- Exercise 17.1 | Set 1

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      Prove that:Question 1. sin2 72o – sin2 60o = (√5 – 1)/8 Solution: We have, L.H.S. = sin2 72o – sin2 60o = sin2 (90o–18o) – sin2 60o = cos2 18o – sin2 60o = [Tex]\left(\frac{\sqrt{10+2\sqrt{5}}}{4}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2[/Tex] = [Tex] \frac{10 + 2\sqrt{5}}{16} – \frac{3}{4}[/Tex]
      4 min read
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