first person will shake hands with 11 other persons. Seconds person will shake hands with 10 other persons, Third person will shake hands with 9 other person and so on.
So , total handshake
= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 66
OR
2 person are to be chosen from 12 person to have a hand shake
? Possible ways = 12C2 = 12! / 2!(12 - 2)! = 66
Number of ways selecting 1 question from ex-7 = 12C1
Number of ways selecting 1 question from ex-8 = 18C1
Number of ways selecting 1 question from ex-9 = 9C1
? Total ways = 12C1 x 18C1 x 9C1 = 12 x 18 x 9 = 1944
Required number of member played will be (139 - 1) = 138
Case I :-
If lady sets on reserved seat, then
2 men can occupy seats from 4 vacant seats in 4P2
= 4 x 3 = 12 ways
Case II :-
If lady does not site on reserved seat, then 1 women can occupy a seat from seat in 4 ways, 1 man can occupy a seat from 3 seats in 3 ways, also 1 man left can occupy a seat from remaining two seats in 2 ways.
? Total ways = 4 x 3 x 2 = 24 ways
Hence, from Case I and case II , total ways = 12 + 24 = 36 ways
Total number of persons = 9
Host can sit in a particular seat in one way .
Now, remaining positions are defined relative to the host .
Hence, the remaining can sit in 8 places in 8P8 = 8! ways.
? The number of required arrangements = 8! x 1 = 8! = 8! ways
Number of 1 digit numbers = 5
Number of 2 digit numbers = 52 = 25
Number of 3 digit numbers = 53 = 125
Number of 4 digit numbers = 54 = 625
Number of 5 digit numbers = 55 = 3125
? Total number of numbers formed with these digits
= 5 + 25 + 125 + 625 + 3125 = 3905
Neglecting the direction of beads in the chain, number of ways of preparing a chain with 5 different coloured beads
= (1/2) x (5 -1)! = (1/2) x 4! = 24/2 = 12
The number of ways of selecting 3 men and 3 women out of 6 men and 5 women = 6C3 x 5C3
= 6!/(3! x 3!) + 5/(3! x 2!)
= 20 + 10 = 30
As, these are two sets of booklets, so number of booklet in each set is 6 and this can be arrange in 6! ways.
Also, the other 6 booklets or 2nd set can also be arranged in other 6 students in 6! ways.
? Required number of ways = 6! x 6!
Total ways = 8C3 x 8C3
= (8 x 7 x 6)/(3 x 2) x (8 x 7 x 6)/(3 x 2)
= 56 x 56 = 3136
3 girls can be be selected out of 5 girls in 5C3 ways. Since, number of boys to be invited is not given, hence out of 4 boys, he can invite them in (2)4 ways.
? Required number of ways 5C3 x (2)4 = 10 x 16 = 160
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