RD Chapter 16 Permutations Ex 16.1 Solutions
Question - 1 : - Compute:
(i) 30!/28!
(ii) (11! тАУ 10!)/9!
(iii) L.C.M. (6!, 7!, 8!)
Answer - 1 : -
(i) 30!/28!
Let us evaluate,
30!/28! = (30 ├Ч 29 ├Ч 28!)/28!
= 30 ├Ч 29
= 870
(ii) (11! тАУ 10!)/9!
Let us evaluate,
We know,
11! = 11 ├Ч 10 ├Ч 9 ├Ч тАж. ├Ч 1
10! = 10 ├Ч 9 ├Ч 8 ├Ч тАж ├Ч 1
9! = 9 ├Ч 8 ├Ч 7 ├Ч тАж ├Ч 1
By using these values we get,
(11! тАУ 10!)/9! = (11 ├Ч 10 ├Ч 9! тАУ 10 ├Ч 9!)/ 9!
= 9! (110 тАУ 10)/9!
= 110 тАУ 10
= 100
(iii) L.C.M. (6!, 7!, 8!)
Let us find the LCM of (6!, 7!, 8!)
We know,
8! = 8 ├Ч 7 ├Ч 6!
7! = 7 ├Ч 6!
6! = 6!
So,
L.C.M. (6!, 7!, 8!) = LCM [8 ├Ч 7 ├Ч 6!, 7 ├Ч 6!, 6!]
= 8 ├Ч 7 ├Ч 6!
= 8!
Question - 2 : - Prove that: 1/9! + 1/10! + 1/11! = 122/11!
Answer - 2 : -
Given:
1/9! + 1/10! + 1/11! = 122/11!
Let us consider LHS: 1/9! + 1/10! + 1/11!
1/9! + 1/10! + 1/11! = 1/9! + 1/(10├Ч9!) + 1/(11├Ч10├Ч9!)
= (110 + 11 + 1)/(11 ├Ч 10 ├Ч 9!)
= 122/11!
= RHS
Hence proved.
Question - 3 : - Find x in each of the following:
(i) 1/4! + 1/5! = x/6!
(ii) x/10! = 1/8! + 1/9!
(iii) 1/6! + 1/7! = x/8!
Answer - 3 : -
(i) 1/4! + 1/5! = x/6!
We know that
5! = 5 ├Ч 4 ├Ч 3 ├Ч 2 ├Ч 1
6! = 6 ├Ч 5 ├Ч 4 ├Ч 3 ├Ч 2 ├Ч 1
So by using these values,
1/4! + 1/5! = x/6!
1/4! + 1/(5├Ч4!) = x/6!
(5 + 1) / (5├Ч4!) = x/6!
6/5! = x/(6├Ч5!)
x = (6 ├Ч 6 ├Ч 5!)/5!
= 36
тИ┤ The value of x is 36.
(ii) x/10! = 1/8! + 1/9!
We know that
10! = 10 ├Ч 9!
9! = 9 ├Ч 8!
So by using these values,
x/10! = 1/8! + 1/9!
x/10! = 1/8! + 1/(9├Ч8!)
x/10! = (9 + 1) / (9├Ч8!)
x/10! = 10/9!
x/(10├Ч9!) = 10/9!
x = (10 ├Ч 10 ├Ч 9!)/9!
= 10 ├Ч 10
= 100
тИ┤ The value of x is 100.
(iii) 1/6! + 1/7! = x/8!
We know that
8! = 8 ├Ч 7 ├Ч 6!
7! = 7 ├Ч 6!
So by using these values,
1/6! + 1/7! = x/8!
1/6! + 1/(7├Ч6!) = x/8!
(1 + 7)/(7├Ч6!) = x/8!
8/7! = x/8!
8/7! = x/(8├Ч7!)
x = (8 ├Ч 8 ├Ч 7!)/7!
= 8 ├Ч 8
= 64
тИ┤ The value of x is 64.
Question - 4 : - Convert the following products into factorials:
(i) 5 тЛЕ 6 тЛЕ 7 тЛЕ 8 тЛЕ 9 тЛЕ 10
(ii) 3 тЛЕ 6 тЛЕ 9 тЛЕ 12 тЛЕ 15 тЛЕ 18
(iii) (n + 1) (n + 2) (n + 3) тАж(2n)
(iv) 1 тЛЕ 3 тЛЕ 5 тЛЕ 7 тЛЕ 9 тАж (2n тАУ 1)
Answer - 4 : -
(i)┬а5┬атЛЕ┬а6┬атЛЕ┬а7┬атЛЕ┬а8┬атЛЕ┬а9┬атЛЕ┬а10
Let us evaluate
We can write it as:
5┬атЛЕ┬а6┬атЛЕ┬а7┬атЛЕ┬а8┬атЛЕ┬а9┬атЛЕ┬а10 = (1├Ч2├Ч3├Ч4├Ч5├Ч6├Ч7├Ч8├Ч9├Ч10)/(1├Ч2├Ч3├Ч4)
= 10!/4!
(ii)┬а3┬атЛЕ┬а6┬атЛЕ┬а9┬атЛЕ┬а12┬атЛЕ┬а15┬атЛЕ┬а18
Let us evaluate
3┬атЛЕ┬а6┬атЛЕ┬а9┬атЛЕ┬а12┬атЛЕ┬а15┬атЛЕ┬а18 = (3├Ч1) ├Ч (3├Ч2) ├Ч (3├Ч3) ├Ч (3├Ч4) ├Ч(3├Ч5) ├Ч (3├Ч6)
= 36┬а(1├Ч2├Ч3├Ч4├Ч5├Ч6)
= 36┬а(6!)
(iii)┬а(n + 1) (n + 2) (n +3) тАж (2n)
Let us evaluate
(n + 1) (n + 2) (n +3) тАж (2n) = [(1) (2) (3) .. (n) тАж (n + 1) (n + 2) (n + 3) тАж (2n)] / (1) (2) (3).. (n)
= (2n)!/n!
(iv)┬а1┬атЛЕ┬а3┬атЛЕ┬а5┬атЛЕ┬а7┬атЛЕ┬а9 тАж (2n тАУ 1)
Let us evaluate
1┬атЛЕ┬а3┬атЛЕ┬а5┬атЛЕ┬а7┬атЛЕ┬а9 тАж (2n тАУ 1) = [(1) (3) (5) тАж (2n-1)][(2) (4) (6) тАж (2n)] / [(2) (4) (6) тАж (2n)]
= [(1) (2) (3) (4) тАж(2n-1) (2n)] / 2n┬а[(1) (2) (3) тАж (n)]
= (2n)! / 2n┬аn!
Question - 5 : - Which of the following are true:
(i) (2 + 3)! = 2! + 3!
(ii) (2 ├Ч 3)! = 2! ├Ч 3!
Answer - 5 : -
(i) (2 + 3)! = 2! + 3!
Let us consider LHS: (2 + 3)!
(2 + 3)! = 5!
Now RHS,
2! + 3! = (2├Ч1) + (3├Ч2├Ч1)
= 2 + 6
= 8
LHS тЙа RHS
тИ┤ The given expression is false.
(ii) (2 ├Ч 3)! = 2! ├Ч 3!
Let us consider LHS: (2 ├Ч 3)!
(2 ├Ч 3)! = 6!
= 6 ├Ч 5 ├Ч 4 ├Ч 3 ├Ч 2 ├Ч 1
= 720
Now RHS,
2! ├Ч 3! = (2├Ч1) ├Ч (3├Ч2├Ч1)
= 12
LHS тЙа RHS
тИ┤ The given expression is false.
Question - 6 : - Prove that: n! (n + 2) = n! + (n + 1)!
Answer - 6 : -
Given:
n! (n + 2) = n! + (n + 1)!
Let us consider RHS = n! + (n + 1)!
n! + (n + 1)! = n! + (n + 1) (n + 1 тАУ 1)!
= n! + (n + 1)n!
= n!(1 + n + 1)
= n! (n + 2)
= L.H.S
L.H.S = R.H.S
Hence, Proved.