7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.
1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)
You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.
In each of these numbers, 7 is written once. Therefore, 243 times.
2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).
There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.
In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.
3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.
Therefore, the total number of times the digit 7 is written between 1 and 999 is
243 + 54 + 3 = 300
There is 1 way to correctly guess who comes in first, second, and third. There is only one set of first, second and third place winners. You must correctly guess these three people, and there is only one way to do so.
There are 5 slots.
__ __ __ __ __
The first slot must be a four. There are 4 ways to put a four in the first slot.
There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the third slot. etc.
(4)(4)(4)(4)(4) = 1024
Therefore there are 1024 different ways to produce the desired hand of cards.
Since order does not matter it is a combination.
The word AND means multiply.
Given 4 basketball, 3 volleyball, 2 soccer.
We want 2 basketball games and 1 other event. There are 5 choices left.
C(n,r)
C(How many do you have, How many do you want)
C(have 4 basketball, want 2 basketball) x C(have 5 choices left, want 1)
C(4,2) x C(5,1) = (6)(5) = 30
Therefore there are 30 different ways in which you can go to two basketball games and one of the other events.
When a hand of cards is dealt, the order of the cards does not matter. If you are dealt two kings, it does not matter if the two kings came with the first two cards or the last two cards. Thus cards are combinations. There are 52 cards in a deck and we want to know how many different ways we can put them in groups of five at a time when order does not matter. The combination formula is used.
C(52,5) = 2,598,960
Since it does not matter what order the committee members are chosen in, the combination formula is used.
Committees are always a combination unless the problem states that someone like a president has higher hierarchy over another person. If the committee is ordered, then it is a permutation.
C(17,7)= 19,448
There are 8 students and the maximum capacity of the cars together is 9.
We may divide the 8 students as follows
Case I: 5 students in the first car and 3 in the second Or
Case II: 4 students in the first car and 4 in the second
Hence, in Case I: 8 students are divided into groups of 5 and 3 in ways.
Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in ways.
Therefore, the total number of ways in which 8 students can travel is
= 56 + 70 = 126.
Since two of the eight first class petty officers are to fill two different offices, we write 56
Then, two of the six second class petty officers are to fill two different offices; thus, we write =30
The principle of choice holds in this case; therefore, the members have 56 x 30 = 1680 ways to select the required office holders
fix one person and the brothers B1 P B2 = 2 ways to do so.
other 17 people= 17!
Each person out of 18 can be fixed between the two=18, thus, 2 x 17! x 18=2 x 18!
Total number of alphabets = 10
so ways to arrange them = 10!
Then there will be duplicates because 1st S is no different than 2nd S.
we have 4 Is 3 S and 2 Ps
Hence number of arrangements = 10!/4! x 3! x 2! = 12600
Selecting zero'A's= 1
Selecting one 'A's = 1
Selecting two 'A's = 1
Selecting three 'A's = 1
Selecting four 'A's = 1
Selecting five 'A's = 1
=> Required number ofways =6
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