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4 months ago

Permutations and Combinations Class 11th

27. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5blue balls if each selection consists of 3 balls of each colour.

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Payal Gupta

Contributor-Level 10

27. Since we are to select 3 balls from each colour in order to select 9 balls from the collections of 6, 5 and 5 balls of red, white and blue colours respectively, we can have the combination

⁶C₃ (red) *⁵C₃ (white) *⁵C₃ (blue)

= $\frac{6!}{3! (6 - 3)!}$ * $\frac{5!}{3! (5 - 3)!}$ * $\frac{5!}{3! (5 - 3)!}$

= 20 * 10 * 10

= 2000

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Permutations and Combinations Class 11th

26. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

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Payal Gupta

Contributor-Level 10

26.The number of ways of selecting a team consisting of 3 boys from 5 boys and 3 girls from 4 girls is

⁵C₃*⁴C₃

= $\frac{5!}{3! (5 - 3)!}$ * $\frac{4!}{3! (4 - 3)!}$

= $\frac{20}{2}$ * $\frac{4}{1}$

= 40

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Permutations and Combinations Class 11th

25. How many chords can be drawn through 21 points on a circle?

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Payal Gupta

Contributor-Level 10

25. A chord is drawn by connecting 2 points on a circle.

As we are given with 21 points on the circle, we have the following combination to find the number of chords.

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Permutations and Combinations Class 11th

24. Determine n if (i) ²ⁿC₃ : ⁿC₃ = 12 : 1 (ii) ²ⁿC₃ : ⁿC₃ = 11 : 1

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Payal Gupta

Contributor-Level 10

24. i. ²ⁿC₃ : ⁿC₃ = 12 : 1

=> $\frac{2 n!}{3! (2 n - 3)!}$ ÷ $\frac{n!}{3! (n - 3)!}$ = $\frac{12}{1}$

=> $\frac{2 n (2 n - 1) (2 n - 2) (2 n - 3)!}{3! (2 n - 3)!}$ * $\frac{3! (n - 3)!}{n (n - 1) (n - 2) (n - 3)!}$ = 12

=> $\frac{2 n (2 n - 1) (2 n - 2)}{n (n - 1) (n - 2)}$ = 12

=> $\frac{2 (2 n - 1) 2 (n - 1)}{(n - 1) (n - 2)}$ = 12

=> 4(2n - 1) = 12(n – 2)

=> 8n – 4 = 12n – 24

=> 24 – 4 = 12n – 8n

=> 20 = 4n

=>n = $\frac{20}{4}$

=>n = 5

ii. ²ⁿC₃ : ⁿC₃ = 11 : 1

=> $\frac{2 n!}{3! (2 n - 3)!}$ ÷ $\frac{n!}{3! (n - 3)!}$ = $\frac{11}{1}$

=> $\frac{2 n (2 n - 1) (2 n - 2) (2 n - 3)!}{3! (2 n - 3)!}$ * $\frac{3! (n - 3)!}{n (n - 1) (n - 2) (n - 3)!}$ = 11

=> $\frac{2 n (2 n - 1) 2 (n - 1)}{n (n - 1) (n - 2)}$ = 11

=> 4(2n – 1) = 11(n – 2)

=> 8n – 4 = 11n – 22

=> 22 – 4 = 11n – 8n

=> 18 = 3n

=>n = $\frac{18}{3}$

=>n = 6

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4 months ago

Permutations and Combinations Class 11th

Class 11 Permutation and Combinations Exercise 6.4 Solutions

23. ⁿC₈ = ⁿC₂, find ⁿC₂.

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Payal Gupta

Contributor-Level 10

23. ⁿC₈ = ⁿC₂

As, ⁿC_a₌ⁿC_b

=>a = b or a = n – b

=>n = a + b

We have,

ⁿC₈ = ⁿC₂

=>n = 8 + 2

=>n = 10

Therefore,

ⁿC₂

= ⁿC₂

= $\frac{10!}{2! (10 ? 2)!}$

= $\frac{10!}{2! 8!}$

= 45

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4 months ago

Permutations and Combinations Class 11th

22. In how many ways can the letters of the word PERMUTATIONS be arranged if the

(i) Words start with P and end with S,

(ii) Vowels are all together,

(iii) Thereare always 4 letters between P and S?

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Payal Gupta

Contributor-Level 10

22. There are 12 letters in which T appears 2 times and rest are all different.

i. When P and S are fixed as first and last letter we can arrange the remaining 10 letter taking all at a time. i.e.

Number of permutation = $\frac{10!}{2!}$

= 18,14,400

ii. We take the 5 vowels (E, U, A, I, O) as one single object. This single object with the remaining 7 object are treated as 8 object which have 2 – T's.

So, number of permutations in which the vowels come together

= permutation of 8 object x permutation within the vowels

= $\frac{8!}{2!}$ * 5!

= 20160 * 120

= 2419200

iii. In order to have 4 letters between P and S, (P, S) should have the possible sets of places (

...more

22. There are 12 letters in which T appears 2 times and rest are all different.

i. When P and S are fixed as first and last letter we can arrange the remaining 10 letter taking all at a time. i.e.

Number of permutation = $\frac{10!}{2!}$

= 18,14,400

ii. We take the 5 vowels (E, U, A, I, O) as one single object. This single object with the remaining 7 object are treated as 8 object which have 2 – T's.

So, number of permutations in which the vowels come together

= permutation of 8 object x permutation within the vowels

= $\frac{8!}{2!}$ * 5!

= 20160 * 120

= 2419200

iii. In order to have 4 letters between P and S, (P, S) should have the possible sets of places (1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11) and (7, 12) similarly (S, P) should have the same possible sets of places.

So, total number of possible positions fixed for P and S = 14

After fixing P and S we can rearrange the remaining 10 places with the remaining 10 letters of which 2 are T's.

So, required permutation = $\frac{10!}{2!}$

= 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3

= 18,14,400

Therefore, total permutation for which there are always 4 letters between P and S

= 14 * 18,14,400

= 2,54,01,600

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4 months ago

Permutations and Combinations Class 11th

21. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?

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Payal Gupta

Contributor-Level 10

21. There are 11 letters of which M appears 1 time, I appears 4 times, S appears 4 times and P appears 2 times.

The required number of arrangements = $\frac{11!}{4! 4! 2!}$

= 11 * 10 * 9 * 5

= 34650

When the four I occurs together we treat them as single object IIII. This single object together with 7 remaining object will account for 8 object which have 1-M. 2-P and 4-S.

So, required number of permutation = $\frac{8!}{4! 2!}$

= 840

Therefore, total no. of permutation in which 4-I's do not come together

= 34650 – 840

= 33810

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4 months ago

Permutations and Combinations Class 11th

20. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.

(i) 4 letters are used at a time,

(ii) all letters are used at a time,

(iii) all letters are used but first letter is a vowel?

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Payal Gupta

Contributor-Level 10

20. i. The permutation of 6 letters in MONDAY taken 4 at a time without repetition is

ii. The permutation of 6 letters in MONDAY when all letters are taken at a time is

⁶P₆ = $\frac{6!}{(6 - 6)!}$ = $\frac{6!}{0!}$ = 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720

iii. The permutation of having one of the two vowels (O, A) as first letter from the word MONDAY when all letters are taken at a time is

²P₁ = $\frac{2!}{(2 - 1)!}$ = 2

After fixing one of the vowel as first letter we can rearrange the remaining 5 letters taking 5 at a time

⁵P₅ = $\frac{5!}{(5 - 5)!}$ = $\frac{5!}{0!}$ = 5! = 5 * 4 * 3 * 2 * 1 = 120

Therefore, total permutation when all letters are used but first letter is vowel from the word MONDAY = 2

...more

20. i. The permutation of 6 letters in MONDAY taken 4 at a time without repetition is

ii. The permutation of 6 letters in MONDAY when all letters are taken at a time is

⁶P₆ = $\frac{6!}{(6 - 6)!}$ = $\frac{6!}{0!}$ = 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720

iii. The permutation of having one of the two vowels (O, A) as first letter from the word MONDAY when all letters are taken at a time is

²P₁ = $\frac{2!}{(2 - 1)!}$ = 2

After fixing one of the vowel as first letter we can rearrange the remaining 5 letters taking 5 at a time

⁵P₅ = $\frac{5!}{(5 - 5)!}$ = $\frac{5!}{0!}$ = 5! = 5 * 4 * 3 * 2 * 1 = 120

Therefore, total permutation when all letters are used but first letter is vowel from the word MONDAY = 2 * 120 = 240

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New question posted

4 months ago

Permutations and Combinations Class 11th

20. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.

(i) 4 letters are used at a time,

(ii) all letters are used at a time,

(iii) all letters are used but first letter is a vowel?

0 Follower 9 Views

New question posted

4 months ago

Permutations and Combinations Class 11th

20. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if 4 letters are used at a time, all letters are used at a time, all letters are used but first letter is a vowel?

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