Explanation
If the best and the worst appear always together, the number of ways 5! × 2. Therefore, required number of ways are 6! – 5! × 2 = 480.
Starting with the first rings. Number of ways it can be worn are 4 as. Now for the second ring it can also be worn in any of the four figures so number of ways are 4 and this is similar for all the six rings. So number of ways are 4 × 4 × 4 × 4 × 4 × 4 × 4 = 46 .
Since the man can go in 4 ways and can back in 3 ways.
Therefore total number of ways are 4 × 3 = 12 ways.
Required number of ways are nmn since each letter may be posted in n ways.
Required number of ways are 210 = 1024, because every question may be answered in 2 ways.
The number will be even if last digit is 2,4,6 or 8 i.e., the last digit can be filled in 4 ways and remaining two digits can be filled in 8P2 ways. Hence required number of numbers is 8P2 × 4 = 224.
The number of possible outcomes with 2 on at least one die = (The total number of outcomes) – (The number of outcomes in which 2 does not appear on any die )
64 – 54 = 1296 – 625 = 671.
Required number of ways are 45 = 1024.
[Since each prize can be distributed in 4 ways]
Required number of ways = 5P3 = 60.
Aliter : For first passenger there are 5 ways to sit. Now for the second passenger, there will be 4 ways and for the third there will be 3 ways to sit. So, total number of ways = 5 × 4 × 3 = 20 × 3 = 60.
It is obvious
Required sum = 3! (3 + 4 + 5 + 6) = 6 × 18 = 108.
[If we fix 3 of the unit place, other three digits can be arranged in 3! ways similarly for 4,5,6.]
Required number of ways = 5! – 4! – 3!
= 120 – 24 – 6 = 90
[Number will be less than 56000 only if either 4 occurs on the first place or 5,4 occurs on the first two places].
Required number of ways = 7P4 = P(7,4).
Sum of the digits in the unit place is 6(2 + 4 + 6 + 8) = 120 units. Similarly, sum of digits in ten place is 120 tens and in hundredth place is 120 hundred, etc. Sum of all the 24 numbers is
120 (1 + 10 + 102 + 103) = 120 × 1111 = 133320
The man can go in 5 ways and he can return in 5 ways . Hence, total number of ways are 5 × 5 = 25.
First prize can be given in 5 ways. Then second prize can be given in 4 ways and the third prize in ways .(Since a competitor cannot get two prizes) and hence the number of ways = 5 × 4 × 3 = 60 ways.
Extreme left place can be filled in 6 ways, the middle place can be filled in 6 ways and extreme right place in only 3 ways.
[Q Number to be formed is odd]
∴ Required number of numbers is = 6 × 6 × 3 = 108.
Numbers are 2, 0, 4, 3, 8
Numbers can be formed
= (Total) – (Those beginning with 0)
= 5! – 4! = 120 – 24 = 96.
Aliter : Digit on the extreme left can be choosen by 4 ways as zero has to be excluded. The second place can be filled by 4 ways and the next place by 3 ways and this process goes on until unit place comes.
Hence , total number of ways = 4 × 4 × 3 × 2 ×1= 96.
Since 12P3 = 1320 ∴ r = 3.
Digit at the extreme left can be choosen by 9 ways as zero cannot be the first digit. Now for the second digit it can be done in 9 ways as consecutive digits are not same. And this is same for next digits. Hence number of ways are
9 × 9 × 9 × ………..n times = 9n.
Number of 1 digit numbers = 6P1
Number of 2 digit numbers = 6P2
Number of 3 digit numbers = 6P3
The required number of numbers
= 6 + 30 + 120 = 156.
Since C and Y are fixed now remaining letters are 6 that can be arranged in 6! ways.
Since L is fixed now 4 letter can be arranged in 4! = 24 ways.
Three letters can be posted in 4 letter boxes in 43 = 64 ways but it consists the 4 ways that all letters may be posted in same box. Hence required ways = 60.
Required number of ways
= 6P3 – 5P2 = 120 – 20 = 100
[Since between 99 and 1000, the numbers are of 3 digits, but those numbers should be omit in which ‘0’ comes at hundred place]
Aliter : Number lying between 99 and 1000 must be of three digits. Hence the first place can be filled by 5 ways excluding zero. As the repetition is not allowed the second place is filled by 5 ways and the last place can be filled by 4 ways. Hence total number of ways are 5 × 5 × 4 = 100.
In forming even numbers , the position on the right can be filled either 0 or 2. When 0 is filled , the remaining positions can be filled in 3! ways and when 2 is filled, the position on the left can be filled in 2 ways (0 cannot be used) and the middle two positions in 2! ways (0 can be used) .Therefore, the number of even numbers formed = 3! + 2(2!) = 10
On simplification you get required result.
The digit at the extreme right can be filled by 8 ways (as 0 and 5 have to be excluded). The digit at the extreme left can be filled by 9 ways and the next two digits can be each filled by 10 ways.
Hence total number of ways are 9 × 10 × 10 × 8 = 7200
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