RD Chapter 16 Permutations Ex 16.3 Solutions
Question - 11 : - If P(n, 5) : P(n, 3) = 2 : 1, find n.
Answer - 11 : -
Given:
P(n, 5) : P(n, 3) = 2: 1
P(n, 5) / P(n, 3) = 2/ 1
By using the formula,
P (n, r) = n!/(n – r)!
P (n, 5) = n!/ (n –5)!
P (n, 3) = n!/ (n –3)!
So, from the question,
P (n, 5) / P(n, 3) = 2/ 1
Substituting theobtained values in above expression we get,
[n!/(n – 5)!] / [n!/ (n – 3)!] = 2/1
[n!/(n – 5)!] × [(n – 3)! / n!] = 2/1
(n – 3)! / (n – 5)! =2/1
[(n –3) (n – 3 – 1) (n – 3 – 2)!] / (n – 5)! = 2/1
[(n –3) (n – 4) (n – 5)!] / (n – 5)! = 2/1
(n – 3)(n – 4) = 2
n2 –3n – 4n + 12 = 2
n2 –7n + 12 – 2 = 0
n2 –7n + 10 = 0
n2 –5n – 2n + 10 = 0
n (n – 5) – 2(n – 5) =0
(n – 5) (n – 2) = 0
n = 5 or 2
For, P (n, r): n ≥ r
∴ n =5 [for, P (n, 5)]
Question - 12 : - Prove that:
1. P (1, 1) + 2. P (2, 2) + 3 . P (3, 3) + … + n . P(n, n) = P(n + 1, n + 1) –1.
Answer - 12 : -
By using the formula,
P (n, r) = n!/(n – r)!
P (n, n) = n!/(n – n)!
= n!/0!
= n! [Since, 0! = 1]
Consider LHS:
= 1. P(1, 1) + 2. P(2,2) + 3. P(3, 3) + … + n . P(n, n)
= 1.1! + 2.2! + 3.3!+………+ n.n! [Since, P(n, n) = n!]
= (2! – 1!) + (3! –2!) + (4! – 3!) + ……… + (n! – (n – 1)!) + ((n+1)! – n!)
= 2! – 1! + 3! – 2! +4! – 3! + ……… + n! – (n – 1)! + (n+1)! – n!
= (n + 1)! – 1!
= (n + 1)! – 1[Since, P (n, n) = n!]
= P(n+1, n+1) – 1
= RHS
Hence Proved.
Question - 13 : - If P(15, r – 1) : P(16, r – 2) = 3 : 4, find r.
Answer - 13 : -
Given:
P(15, r – 1) : P(16, r– 2) = 3 : 4
P(15, r – 1) / P(16, r– 2) = 3/4
By using the formula,
P (n, r) = n!/(n – r)!
P (15, r – 1) = 15! /(15 – r + 1)!
= 15! / (16 – r)!
P (16, r – 2) =16!/(16 – r + 2)!
= 16!/(18 – r)!
So, from the question,
P(15, r – 1) / P(16, r– 2) = 3/4
Substituting theobtained values in above expression we get,
[15!/ (16 – r)!] / [16!/(18 – r)!] = 3/4
[15!/ (16 – r)!] × [(18 – r)! / 16!] = 3/4
[15!/ (16 – r)!] × [(18 – r) (18 – r – 1) (18 – r – 2)!]/(16×15!) = 3/4
1/(16 – r)! × [(18 –r) (17 – r) (16 – r)!]/16 = 3/4
(18 – r) (17 – r) =3/4 × 16
(18 – r) (17 – r) = 12
306 – 18r – 17r + r2 =12
306 – 12 – 35r + r2 =0
r2 –35r + 294 = 0
r2 –21r – 14r + 294 = 0
r(r – 21) – 14(r – 21)= 0
(r – 14) (r – 21) = 0
r = 14 or 21
For, P(n, r): r ≤ n
∴ r =14 [for, P(15, r – 1)]
Question - 14 : - n+5Pn+1 = 11(n – 1)/2 n+3Pn,find n.
Answer - 14 : -
Given:
n+5Pn+1 = 11(n – 1)/2 n+3Pn
P (n +5, n + 1) = 11(n– 1)/2 P(n + 3, n)
By using the formula,
P (n, r) = n!/(n – r)!
(n + 5) (n + 4) = 22(n – 1)
n2 +4n + 5n + 20 = 22n – 22
n2 +9n + 20 – 22n + 22 = 0
n2 –13n + 42 = 0
n2 –6n – 7n + 42 = 0
n(n – 6) – 7(n – 6) =0
(n – 7) (n – 6) = 0
n = 7 or 6
∴ The value of ncan either be 6 or 7.
Question - 15 : - In how many ways can five children stand in a queue?
Answer - 15 : -
Number of arrangementsof ‘n’ things taken all at a time = P (n, n)
So by using theformula,
By using the formula,
P (n, r) = n!/(n – r)!
The total number ofways in which five children can stand in a queue = the number of arrangementsof 5 things taken all at a time = P (5, 5)
So,
P (5, 5) = 5!/(5 – 5)!
= 5!/0!
= 5! [Since, 0! = 1]
= 5 × 4 × 3 × 2 × 1
= 120
Hence, Number of waysin which five children can stand in a queue are 120.
Question - 16 : - From among the 36 teachers in a school, one principal and onevice-principal are to be appointed. In how many ways can this be done?
Answer - 16 : -
Given:
The total number ofteachers in a school = 36
We know, number ofarrangements of n things taken r at a time = P(n, r)
By using the formula,
P (n, r) = n!/(n – r)!
∴ The total numberof ways in which this can be done = the number of arrangements of 36 thingstaken 2 at a time = P(36, 2)
P (36, 2) = 36!/(36 –2)!
= 36!/34!
= (36 × 35 × 34!)/34!
= 36 × 35
= 1260
Hence, Number of waysin which one principal and one vice-principal are to be appointed out of total36 teachers in school are 1260.