Question

The number of permutation from the letter A to G so that neither the set BEG nor CAD appears is
a) \[\dfrac{7!}{{{\left( 3! \right)}^{2}}}\]
b) 3! × (3)2 × 89
c) $\dfrac{7!}{{{\left( 3! \right)}^{3}}}$
d) None

Answer 1

Hint: We see here that, the total number of permutation for 'A' to 'G' is 7! as because there are 7 digits in between 'A' to 'G'.Now if neither the set 'BEG' occurs nor the set 'CAD' occurs then, we get;
No. of ways in which 'BEG'
occurs=n('BEG')=(7-2)!=5!
No. of ways in which 'CAD' occurs=n('CAD')=(7-2)!=5!
No. of ways for the occurrence of both 'BEG' and 'CAD' = ( occurrence of 'BEG' ∩ occurrence of 'CAD')= 3!
So, occurrence of ('BEG'∪'CAD')=n('BEG') + n('CAD') - n('BEG'∩'CAD').

Complete step by step solution:
Total letters for which we have to find the permutations are A to G; that is:- A, B, C, D, E, F, G.
Total A to G = 7;
So, the total number of permutations from A to G is = 7!
Let ‘L’ denote that ‘BEG’ occurs and let ‘M’ denote that ‘CAD’ occurs.
So, the number of ways in which ‘BEG’ occurs = (7 − 2)! = 5!
and the number of ways in which ‘CAD’ occurs = (7 − 2)! = 5!
And, the number of ways in which both ‘BEG’ and ‘CAD’ occurs is;
Number of permutation when both ‘L’ and ‘M’ occur (L∩M) = 3!
 n(L∪M) = 5! + 5! − 3!
⇒ n(L∪M) = 120 + 120 − 6 [:n(A∪B) = n(A) + n(B) − n(A∩B)
=240 − 6
= 234
Since, $5!=5\times 4\times 3\times 2\times 1=120$
and, $3!=3\times 2\times 1=6$
 Required number of ways = 7! − 234
= 5040 − 234
= 4806.
Now checking the options one-by-one we get;
$4806=3!\ \times \ {{\left( 3 \right)}^{2}}\times 89$

Note: If A and B are two finite sets, then
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Simply, the number of elements in the union of set A and B is equal to the sum of cardinal numbers of the sets A and B, minus that of their intersection.