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
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!
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
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
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
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
The number of arrangement in which A and B are not together
= Total number of arrangement - Number of arrangement in which A and B are together
= 4! - 3! x 2! = 24 - 12 = 12
Total number of ways filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls = 25 = 32
Two adjacent boxes with blue balls can be obtained in 4 ways, i.e., (12), (23), (34) and (45). Three adjacent boxes with blue balls can be obtained in 3 ways i.e., (123), (234) and (345). Four adjacent boxes with blue balls can be obtained in 2 ways i.e., (1234) and (2345) and five boxes with blue balls can be got in 1 way.
Hence, the total number of ways of filling the boxes such that adjacent boxes have blue balls
= (4 + 3 + 2 + 1)
= 10
Hence, the number of ways of filling up the boxes such that no two adjacent boxes have blue balls
= 32 - 10
= 22
The first marble can be put into the pockets in 4 ways, so can the second and third.
Thus, the number of ways in which the child can put the marbles = 4 x 4 x 4 = 64 ways .
When All S are taken together, then ASSASSINATION has 10 letters .
\
So, 10 letters in total can be arranged in 10 ! ways .
[? All 'S' are considered as 1]
But, here are 3 'A' and 2'I' and 2 'N' .
? Required number of ways = 10 ! / (3 ! x 2 ! x 2 !) = 151200
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I have a which questions we want to divide the total outcomes