Chapter 3 Permutations and Combination Ex 3.6 Solutions

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.6

(b)⁸⁰C₂
Solution:

(c)¹⁵C₄+¹⁵C₅
Solution:

(d)²⁰C₁₆–¹⁹C₁₆
Solution:

(b)²ⁿC₃:ⁿC₂= 52 : 3
Solution:

(c)ⁿC_n-3= 84
Solution:

(b)ⁿC_r-1:ⁿC_r:ⁿC_r+1= 20 : 35 : 42
Solution:

(b) When 4 of them are collinear.
If no three points are collinear, we get a total of¹⁰C₂= 45 lines by joining them. …..[From (i)]
Since 4 points are collinear, only one line passes through these points instead of⁴C₂lines.
∴⁴C₂– 1 extra lines are included in 45 lines.
Number of lines passing through these points
= 45 – (⁴C₂– 1)
= 45 – + 1
= 45 – + 1
= 45 – 6 + 1
= 40

(b) When 4 of these points are collinear.
If no three points are collinear, total we get¹²C₃= 220 triangles by joining them. ……[From (i)]
Since 4 points are collinear, no triangle can be formed by joining these four points.
∴⁴C₃extra triangles are included in 220 triangles.
∴ Number of triangles that can be obtained from these points =¹²C₃–⁴C₃
= 220 –
= 220 –
= 220 – 4
= 216

(ii)²³C_3n=²³C_2n+3
Solution:
²³C_3n=²³C_2n+3
IfⁿC_x=ⁿC_y, then either x = y or x = n – y
∴ 3n = 2n + 3 or 3n = 23 – 2n – 3
∴ n = 3 or n = 4

(iii)²¹C_6n=
Solution:
²¹C_6n=
IfⁿC_x=ⁿC_y, then either x = y or x = n – y
∴ 6n = n²+ 5 or 6n = 21 – (n²+ 5)
∴ n²– 6n + 5 = 0 or 6n = 21 – n²– 5
∴ n²– 6n + 5 = 0 or n²+ 6n – 16 = 0
If n²– 6n + 5 = 0, then (n – 1)(n – 5) = 0
∴ n = 1 or n = 5
If n = 5 then
n²+ 5 = 30 > 21
∴ n ≠ 5
∴ n = 1
If n²+ 6n – 16 = 0, then (n + 8)(n – 2) = 0
n = -8 or n = 2
n ≠ -8
∴ n = 2
∴ n = 1 or n = 2

Check:
n = 2
∴ n²+ 5 = 2²+ 5 = 9
²¹C_6n=²¹C₁₂
and =²¹C₉
∴ⁿC_r=ⁿC_n-r
∴²¹C₁₂=²¹C₉
∴ n = 2 is a right answer.

(iv)²ⁿC_r-1=²ⁿC_r+1
Solution:
²ⁿC_r-1=²ⁿC_r+1
IfⁿC_x=ⁿC_y, then either x = y or x = n – y
∴ r – 1 = r + 1 or r – 1 = 2n – (r + 1)
But r – 1 = r + 1 is not possible
∴ r – 1 = 2n – (r + 1)
∴ r + r = 2n
∴ r = n

Check:
²ⁿC_r-1=²ⁿC_n-1
and²ⁿC_r+1=²ⁿC_n+1
usingⁿC_r=ⁿC_n-r, we have
²ⁿC_n+1=²ⁿC_2n-(n+1)=²ⁿC_n-1
∴²ⁿC_r-1=²ⁿC_r+1

(v)ⁿC_n-2= 15
Solution:
ⁿC_n-2= 15
∴ⁿC₂= 15 …..[∵ⁿC_r=ⁿC_n-r]
∴
∴
∴ n(n – 1) = 30
∴ n(n – 1) = 6 × 5
Equating both sides, we get
∴ n = 6

(b)¹³C_rand⁸C_r
Solution:
Greatest value of¹³C_r.
Here n = 13, which is odd.
Greatest value of nCr occurs at r = if n is odd.
∴ r =
∴ r = = 6

∴ Difference between the greatest values of¹³C_rand⁸C_r=¹³C_r–⁸C_r
= 1716 – 70
= 1646

(c)¹⁵C_rand¹¹C_r
Solution:
Greatest value of¹⁵C_r.
Here n = 15, which is odd.
Greatest value of nCr occurs at r = if n is odd.
∴ r =
∴ r = = 7

∴ Difference between the greatest values of¹⁵C_rand¹¹C_r=¹⁵C_r–¹¹C_r
= 6435 – 462
= 5973

(ii) Number of committees with men in majority = Total number of committees – (Number of committees with women in majority + women and men equal in number)
=¹⁸C₁₀– 36873
=¹⁸C₈– 36873
= 43758 – 36873
= 6885

(b) When 2 specified students are not included, then 5 students can be selected from the remaining (11 – 2) = 9 students.
∴ Number of ways of selecting 5 students from 9 students =⁹C₅
=
=
= 126
∴ Selection of students is done in 126 ways when 2 specified students are not included.