Let the beds be numbered 1 to 7.
Case 1 : Suppose Anju is allotted bed number 1.
Then, Parvin cannot be allotted bed number 2.
So Parvin can be allotted a bed in 5 ways.
After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.
So, in this case the beds can be allotted in 5´5!ways = 600 ways.
Case 2 : Anju is allotted bed number 7.
Then, Parvin cannot be allotted bed number 6
As in Case 1, the beds can be allotted in 600 ways.
Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6.
Parvin cannot be allotted the beds on the right hand side and left hand side of Anju?s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.
Therefore, Parvin can be allotted a bed in 4 ways in all these cases.
After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.
Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways.
The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways
Total number of letters in the word ABYSMAL are 7
Number of ways these 7 letters can be arranged are 7! ways
But the letter is repeated and this can be arranged in 2! ways
Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.
The 7 letters word 'POVERTY' be arranged in ways = 7! = 5040 ways.
Let 4 girls be one unit and now there are 6 units in all.
They can be arranged in 6! ways.
In each of these arrangements 4 girls can be arranged in 4! ways.
Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280
Given are the two AP'S:
15,12,9.... in which a=15, d=-3.............(1)
-15,-13,-11..... in which a'=-15 ,d'=2.....(2)
now using the nth term's formula,we get
a+(n-1)d = a'+(n-1)d'
substituting the value obtained in eq. 1 and 2,
15+(n-1) x (-3) = -15+(n-1) x 2
=> 15 - 3n + 3 = -15 + 2n - 2
=> 12 - 3n = -17 + 2n
=> 12+17 = 2n+3n
=> 29=5n
=> n= 29/5
Possibilities Bowlers Batsmen Number of ways
6 9
1 4 7
2 5 6
3 6 5
= 15 x 36 = 540
= 6 x 84 = 504
= 1 x 126 = 126
Total = 1170
The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.
The first 24 of these words will start with A.
Then the 25th word will start will CA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.
The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.
Therefore, the rank of CHASM will be 24+6+2= 32
Let last digit is 2
when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways.
so for last digit = 2, total numbers=240
Similarly for 4 and 6
When last digit = 4, total no. of ways =240
and last digit = 6, total no. of ways =240
so total of 720 even numbers are possible.
As the toys are distinct and not identical,
For each of the 8 toys, we have three choices as to which child will receive the toy. Therefore, there are ways to distribute the toys.
Hence, it is and not .
Given word is THERAPY.
Number of letters in the given word = 7
Number of vowels in the given word = 2 = A & E
Required number of different ways, the letters of the word THERAPY arranged such that vowels always come together is
6! x 2! = 720 x 2 = 1440.
Here the order of choosing the elements doesn?t matter and this is a problem in combinations.
We have to find the number of ways of choosing 4 elements of this set which has 11 elements.
This can be done in 11 ways = 330 ways
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