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
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
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
Total number of ways of 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 can be obtained in 4 ways, i.e., (12), (23), (34) and (45).
Three adjacent boxes with blue can be obtained in 3 ways, i.e., (123), (234)and (345). Four boxes with blue can be obtained in 2 ways, i.e., (1234) and (2345). And five boxes with blue can be got in 1 way. Hence, the number of ways of filling the boxes such that adjacent boxes have blue
= (4 + 3 + 2 + 1) = 10.
Hence, the number or ways of filling up the boxes such that no two adjacent boxes have blue = 32 - 10 = 22.
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