GSEB Solutions Class 11 Statistics Chapter 6 Permutations, Combinations and Binomial Expansion Ex 6.1

   

Gujarat Board Statistics Class 11 GSEB Solutions Chapter 6 Permutations, Combinations and Binomial Expansion Ex 6.1 Textbook Exercise Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Statistics Chapter 6 Permutations, Combinations and Binomial Expansion Ex 6.1

Question 1.
Obtain the values of the following:
(1) 10P2
Answer:
10P2 = \(\frac{10 !}{(10-3) !}\)
= \(\frac{10 !}{7 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 !}{7 !}\)
= 10 × 9 × 8 = 720
10P2 = 720

Alternative method:
10P2 = 10(10 – 1)(10 – 2)
= 10 × 9 × 8 = 720
10P2 = 720

(2) 50P2
Answer:
50P2 = \(\frac{50 !}{(50-2) !}\)
= \(\frac{50 !}{48 !}=\frac{50 \times 49 \times 48 !}{48 !}\) = 50 × 49 = 2450
50P2 = 2450

Alternative method:
50P2 = 50 (50 – 1)
= 50 × 49 = 2450
50P2 = 2450

(3) 8P7
Answer:
8P7 = \(\frac{8 !}{(8-7) !}\)
= \(\frac{8 !}{1 !}\)
= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1= 40320
8P7 = 40320

Alternative method:
8P7 = 8 (8 – 1) (8 – 2) (8 – 3) (8 – 4) (8 – 5) (8 – 6) .
= 8 × 7 × 6 × 5 × 4 × 3 × 2 = 40320
8P7 = 40320

(4) 9P9
9P9 = \(\frac{9 !}{(9-9) !}\)
= \(\frac{9 !}{0 !}\)
= 9!
= 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362880
9P9 = 362880

GSEB Solutions Class 11 Statistics Chapter 6 Permutations, Combinations and Binomial Expansion 6.1

Question 2.
If nP3 = 990, then find the value of n.
Answer:
nP3 = 990
∴ n(n – 1)(n – 2)
∴ n(n – 1)(n – 2)= 11 × 10 × 9
∴ n (n – 1) (n – 2) = 11(11 – 1) (11 – 2)
∴ n = 11

Question 3.
If 9Pr = 3024, find the value of r.
Answer:
nPr = \(\frac{n !}{(n-r) !}\)
9Pr = \(\frac{9 !}{(9-r) !}\)
∴ 3024 = \(\frac{362880}{(9-r) !}\)
∴ (9 – r)! = \(\frac{362880}{3024}\)
∴ (9 – r)! = 120 = 5!
∴ 9 – r = 5
∴ r = 9 – 5 = 4
Hence, r = 4

Question 4.
If 3.(n + 3)p4 = 5.(n + 2)p4 then find the value of n.
Answer:
3.(n + 3)p4 = 5.(n + 2)p4
∴ 3.(n + 3) (n + 3 – 1) (n + 3 – 2) (n + 3 – 3) [According to definition of nPr]
= 5.(n + 2) (n + 2 – 1) (n + 2 – 2) (n + 2 – 3) 3.(n + 3) (n + 2) (n + 1) (n)
= 5.(n + 2) (n + 1) (n) (n- 1)
∴ 3(n + 3) = 5(n – 1)
∴ 3n + 9 = 5n – 5
∴ 9 + 5 = 5n – 3n
∴ 14 = 2n
∴ n = \(\frac{14}{2}\) = 7

Question 5.
In how many ways can 4 persons be arranged in a row?
Answer:
4 persons are to be arranged in a row.
∴ n = 4, r = 4
∴ Total permutations = 4P4
= 4! (∵ nPn = n !)
= 4 × 3 × 2 × 1
= 24

Question 6.
How many six digit numbers can be formed using all the digits 1, 2, 3, 0, 7, 9?
Answer:
Using all the digits 1, 2, 3, 0, 7, 9, six digit numbers are to be formed.
∴ Excluding digit 0, one of the five digits can be placed at the first place in 5P1 ways

Now, remaining 5 digits (including 0) can be arranged in remaining 5 places in 5P5 ways.

∴ Total permutations for six digit numbers
= 5P1 × 5P5
= 5 × 5!
= 5 × 120
= 600
Hence, 600 numbers of 6 digit can be formed.

Question 7.
In how many ways can 5 boys and 3 girls be arranged In a row such that all the boys are together?
Answer:
5 boys and 3 gIrls are to be arranged In a row such that all the boys are together.
GSEB Solutions Class 11 Statistics Chapter 6 Permutations, Combinations and Binomial Expansion 6.1 1
5 boys are to be arranged together. therefore considering them as one person total 4 persons can be arranged In 4P4 ways.

Now. In each of these arrangments 5 boys can be arranged among themselves in 5P5 ways.
∴ Total permutation = 4P4 × 5P5
= 4! × 5!
= 24 × 120
= 2880
Hence, 5 boys and 3 gIrls can be arranged In a row such that 5 boys are together in 2880 ways.

Question 8.
There are 7 cages for 7 lions in a zoo. 3 cages out of 7 cages are so small that 3 out of 7 lions cannot fit in it. In bow many ways can 7 lions be caged in 7 cages?
GSEB Solutions Class 11 Statistics Chapter 6 Permutations, Combinations and Binomial Expansion 6.1 2
Answer:
Except 3 small cages In remaining 4 cages 3 lions can be caged In 4P3 ways.

Now, In the remainIng (1 big + 3 small) 4 cages the remainIng 4 lions can be caged In 4P4 ways.
∴ Total permutations of cagIng 7 lions in
7 cages = 4P3 . 4P4
= (4 × 3 × 2) × 4!
= 24 × 24
= 576

Question 9.
Using all the digIts 2, 3, 5, 8, 9, how many numbers greater than 50,000 can be formed?
Answer:
Given digits are 2, 3, 5, 8, 9.
For the numbers greater than 50.000 the digit at the first place may be 5 or greater than It.
∴ Out of the digit 5, 8, 9, the first digit can be placed in 3P1 ways.

Now, from the remaIning 4 dIgits, all four can be placed in 4P4 ways.
∴ Total permutations for the numbers greater than 50,000 = 3P1 × 4P4
= 3 × 4!
= 3 × 24
= 72

GSEB Solutions Class 11 Statistics Chapter 6 Permutations, Combinations and Binomial Expansion 6.1

Question 10.
A person has 5 chocolates of different sizes. These chocolates are to be distributed among 5 children of different ages. If the biggest chocolate Is to be given to the youngest child then In how many ways, 5 chocolates can be distributed among 5 children?
Answer:
A person has 5 chocolates of different sizes and they are to be distributes among 5 children of different ages. If the biggest chocolate is to be given to the youngest child, then remaining 4 chocolates can be distributed among the remaining 4 children in 4P4 ways.
∴ Total permutations = 1P1 x 4P4
= 1 × 4!
= 1 × 24
= 24

Question 11.
How many total arrangements can be made using all letters of the following Words?
(1) STATISTICS
Answer:
In this word there are lo letters of which I is repeated 2 times. S is repeated 3 times and T is repeated 3 times.
Here, n = 10: p = 2; q = 3; r = 3
∴ Total permutation
= \(\frac{n !}{p ! q ! r !}\)
= \(\frac{10 !}{2 ! 3 ! 3 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1}\)
= \(\frac{3628800}{72}\) = 50400

(2) BOOKKEEPER
Answer:
According to permutation of Identical things,
Total permutations = \(\frac{n !}{p ! q ! r !}\)
Here, n = 10; p = 3; q = 2; r = 2
∴ Total permutations
= \(\frac{10 !}{3 ! 2 ! 2 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 2 \times 2}\)
= \(\frac{3628800}{24}\) = 151200

(3) APPEARING
Answer:
There are 9 letters in this word of which A is repeated 2 times and P is repeated 2 times.
AccordIng to permutation of identical things.
Total permutations = \(\frac{n !}{p ! q !}\)
Here, n = 9; p = 2; q = 2
∴ Total permutations = \(\frac{9 !}{2 ! 2 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 2}\)
= \(\frac{362880}{4}\) = 90720

Question 12.
What is the ratio of number of arrangements of all letters of the word ASHOK and GEETA ?
Answer:
Total permutations of arranging 5 letters of word ASHOK = 5P5 = 5! = 120

Total permutations of arranging 5 letters of which E is repeated 2 times of word GEETA = \(\frac{5 !}{2 !}=\frac{120}{2}\) = 60
∴ The ratio of number of arrangements of all letters of the word ASHOK and GEETA
= \(\frac{120}{2}\) = 2: 1

Question 13.
There are 5 seats in a car including the driver’s seat. If 3 out of 10 members in a family know driving then in how many ways, 5 persons out of 10 members can be arranged in the car?
Answer:
3 out of 10 members in a family know driving.
∴ On the driver’s seat a member can be place in 3P1 ways.

Now, there are 5 seats in a car including the driver’s seat.
The remaining 4 members out of remaining 9 members of the family can be placed in 9P4 ways.

∴ Total permutations of such arrangement
= 3P1 × 9P4
= 3 × (9 × 8 × 7 × 6)
= 3 × 3024
= 9072

Question 14.
If all the arrangements formed using aU the letters of the following words are arranged In the order of dictionary then what will be the rank of that word?
(1 ) PINTU
Answer:
There are 5 letters I I. N, T, U in the word PINTU.
These 5 letters can be arrange in 5P5 = 5! = 120 ways.

Now, we have to obtain the order of the word PINTU from all 120 arrangements as per thern dictionary order.
Alphabetical order of all the letters of the word PINTU is I, N, P T, U.

Number of words with I at the first place
= 1P1 × 4P4 = 24

Number of words with N at the first place
= 1P1 × 4P4 = 24

Number of words with P at the first place, I at the second place. N at the third place. T at the fourth place and U at the fifth place = 1P1 × 1P1 × 1P1 × 1P1 × 1P1 = 1 and that word is PINTU itself.

∴ Dictionary order of the word PINTU
= 24 + 24 + 1 = 49.

(2) NURI
Answer:
‘There are 4 letters N. U, R, I in the word NURI.
These 4 letters can be arrange In 4P4 = 4! = 24 ways.

Now, we have to obtain the order of the word NURI from all 24 arrangements as per the dictionary order.
Alphabetical order of all the letters of the word NURI is I. N. R, U.

Number of words with I at the first place = 1P1 × 3P3 = 3! = 6
Number of words with N at the first place and I at the second place = 1P1 × 1P1 × 2P2 = 1 × 1 × 2! = 2

Number of words with N at the first place and R at the second place = 1P1 × 1P1 × 2P2 = 1 × 1 × 2! = 2

Number of words with N at the first place, U at the second place and I at the third place = 1P1 × 1P1 × 1P1 × 1P1 = 1.

Now, the word with N at the first place, U at the second place, R at the third place and I at the fourth place is the word NURI itself.
∴ Dictionary order of the word NURI = 6 + 2 + 2 + 1 + 1 = 12

(3) NIRAL
Answer:
There are 5 letters N, I, R, A, L in the word NIRAL.
These 5 letters can be arranged in 5P5 = 5! = 120 ways.

Now, we have to obtain the order of the word NIRAL from all 120 arrangements as per the dictionary order.
The alphabetical order of all letters of the word NIRAL is A, I, L, N, R.

Number of words with A at the first place
= 1P1 × 4P4 = 1 × 4! = 24

Number of words with I at the first place
= 1P1 × 4P4 = 1 × 4! = 24

Number of words with L at the first place
= 1P1 × 4P4 = 1 × 4! = 24

Number of words with N at the first place and A at the second place = 1P1 ×1P1 × 3P3
= 1 × 1 × 3! = 1 × 6 = 6.

Number of words with N at the first place, I at the second place and A at the third place = 1P1 × 1P1 × 1P1 × 2P2 = 1 × 2! = 2

Number of words with N at the first place, I at the second place and L at the third place = 1P1 × 1P1 × 1P1 × 2P2 = 1 × 2! = 2

Now, the words with N at the first place, I at the second place, and R at the third place are NIRAL, NIRLA,
∴ Dictionary order of the word NIRAL = 24+ 24+ 24 + 6 + 2 + 2 + 1 = 83

(4) SUMAN
Answer:
There are 5 letters S, U, M, A, N in the word SUMAN.
These 5 letters can be arrange in 5P5 = 5! = 120 ways.

Now, we have to obtain the order of the word SUMAN from all 120 arrangements as per the dictionary order.
The alphabetical order of all letters of the word SUMAN is A, M, N, S, U.

Number of words with A at the first place
= 1P1 × 4P4 = 1 × 4! = 24

Number of words with M at the first place
= 1P1 × 4P4 = 1 × 4! = 24

Number of words with N at the first place
= 1P1 × 4P4 = 1 × 4! = 24

Number of words with S at the first place and A at the second place
= 1P1 × 3P3 = 1 × 3! = 6

Number of words with S at the first place and M at the second place
= 1P1 × 3P3 = 1 × 3! = 6

Number of words with S at the place and N at the second place
= 1P1 × 3P3 = 1 × 3! = 6

Number of words with S at the first place, U at the second place and A at the third place = 1P1 × 1P1 × 1P1 × 2P2 = 1 × 2! = 2

Now, the words with S at the first place, U at the second place and M at the third place are SUMAN and SUMNA.
∴ Dictionary order of the word SUMAN is = 24+ 24+ 24 + 6 + 6 + 6 + 2+1 = 93

GSEB Solutions Class 11 Statistics Chapter 6 Permutations, Combinations and Binomial Expansion 6.1

Question 15.
How many arrangements of the letters of the word SHLOKA can be made such that all vowels are together?
Answer:
O and A are two vowels in the word SHLOKA-
GSEB Solutions Class 11 Statistics Chapter 6 Permutations, Combinations and Binomial Expansion 6.1 3
There are 6 letters in the word SHLOKA of which 2 are vowels.
Considering 2 vowels as one letter, total 5 letters can be arranged in P5 ways and in each of these arrangements two vowels can be arranged among themselves in P2 ways.

∴ Total permutations = 5P5 × 3P3 = 5 ! × 2!
= 120 × 2 = 240

Question 16.
7 speakers A, B, C, D, E, F and G are Invited to deliver a speech in a program. Speakers have to deliver speech one after the other. In how many ways, speeches of 7 speakers can be arranged If the speaker B has to deliver his speech Immediately after the speaker A?
Answer:
7 speakers A, B. C. D, E, F. G have to deliver speech one after the other. If the speaker A delivers lits speech first. the remaining 6 speakers can be arranged for their speech in 6P1 ways.

If B delivers his speech after the speaker A. the remaIning 5 speakers can be arranged for their speech In 5P1 ways.
If C delivers his speech after the speakers A and B, the remaining 4 speakers can be arranged for their speech in 4P1 ways.
If D delivers his speech after the speakers A, B and C, the remaIning 3 speakers can be arranged for their speech in 3P1 ways.
If E delivers his speech after the speakers A, B, C and D, the remalng 2 speakers can be arranged for their speech in 2P1 ways.

Not, at the last the speaker F can deliver his speech after the speakers A, B, C, D and E, have delivered their speech in 1P1 ways.

∴ Total permutations
= 6P1 × 5P1 × 4P1 × 3P1 × 2P1 × 1P1
= 6 × 5 × 4 × 3 × 2 × 1
= 720

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