There are 10 station on railway line.
So, the number of different journey tickets between two station from given 10 stations from one side = 10C2 = 10 x 9/2 = 45.
Similarly, number of different journey tickets from other side = 45
? Total number of tickets to be generated by authorities. = 45 + 45 = 90
Here, we have two sections A and B. Section A has 3 question and B has 5 question and one question from each section is compulsory according to the given condition.
? Number of ways selecting one or more than one question from section A
= 23 - 1 = 7
Similarly, from section B = 25 - 1 = 31
According to the rule of multiplication, the required number of ways in which a candidate can select the question
= 7 x 31 = 217
The required number of triangles = nC3 - mC3
Here, n = 14, m = 4
= 14C3 - 4C3
= (14 x 13 x 12 x 11! ) / (3! x 11!) - 4! / (3! x 1!)
= (14 x 13 x 12)/6 - 4/1
= 14 x 26 - 4
= 364 - 4
= 360
First , we select 13 persons out of 24 persons in 24 C13 ways.
Now, these 13 persons can be seated in 12! ways around a table .
So required number of ways = 24 C13 x 12!
= [24! / {13!(24 - 13)!}] x 12!
= [24! / {13! x 11!}] x 12!
= 24! / (13 x 11!)
There is a 7 - digit telephone number but extreme right and extreme left positions are fixed .
i. e. , 6 x x x x x 5
? Required number of ways = 8 x 7 x 6 x 5 x 4 = 6720
Married couples :
MF MF MF MF ? AB, CD, EF, CD
Possible teams :
AD CB EB GB
AF CF ED GD
AH CH EH GF
Team AD can play only with CB, CF, CH, EB, EH, GB, GF (7 teams).
Teams AD cannot play with AF, AH, ED and GD.
The same will apply with all teams, So, number of total matches = 12 x 7 = 84
But every match includes 2 teams, so the actual number of matches = 84/2 = 48
If Mrs. X is selected among the ladies in the committee, then Mr. Y is not selected or if Mrs. X is not selected then Mr. Y can be there in the committee...
So, required number of ways
= 8C3 x 6C4 + 7C3 x 7C4
= [(8 x 7 x 6)/(3 x 2)] x [(6 x 5)/(2 x 1)] + [(7 x 6 x 5)/(3 x 2)] x [(7 x 6 x 5)/(3 x 2)]
= 840 + 1225
= 2065
A five-digit number, which is divisible by 3, is formed when sum of digits is also divisible by 3.
So, combination formed using six-digits, which are divisible by 3
= 5 + 4 + 3 + 2 + 1 = 15
= 5 + 4 + 2 + 1 + 0 = 12
So, set of number are (5, 4, 3, 2, 1) and (5, 4, 2, 1, 0).
Number formed by using 1st set = 5 x 4 x 3 x 2 x 1 = 120
Similarly, using 2nd set = 4 x 4 x 3 x 2 x 1 = 96
Hence, using 2nd set, underlined place cannot be filled by 0, otherwise it will become a four-digit number.
? Total number = 120 + 96 = 216
Any of the 4 colour can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe. The third stripe can again be coloured in 3 ways (we can repeat the colour of the first stripe but not use the colours of the second stripe).
Similarly, There are 3 ways to colour each of the remaining stripes.
? The number of ways the flag can be coloured is 4(3)5 = (12) (3)4 = 12 x 81
The available digits are 0, 1, 2,...., 9. The first digit can be chosen in 9 ways (0 not acceptable), the second digit can be accepted in 9 ways (digit repetition not allowed). Thus, the code can be made in 9 x 9 = 81 ways.
Now, there are only 4 digits which can create confusion 1, 6, 8, 9. The same can be given in the following ways
Total number of ways confusion can arise
= 4 x 3 = 12
Thus, the ways in which no such confusion arise = 81-12 =69
The digit in the unit's place should be greater than that in the tens' place.
Hence, if digit 5 occupies the unit place, then remaining four digits need not to follow any order,hence required number = 4!
However, if digit 4 occupies the unit place then 5 cannot occupy the ten;s position. Hence, digit at the ten's place and it will be filled by the digit 1, 2 or 3. This can happen in 3 ways. The remaining 3 digit can be filled in the remaining three place in 3! ways.
Hence, in all, we have (3 x 3!) numbers ending in 4. Similarly, if we have 3 in the unit's place and it will be either 1 or 2. this can happen in 2 ways. Hence, we will have (2 x 3! ) number ending in 3 . Similarly, we can find that there will be 3! numbers ending in 2 and no number with 1. Hence, total number of numbers
= 4! + (3) x 3! + (2 x 3!) + 3!
= 4! + 6 x 3! = 24 + (6 x 6) = 60
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