hey meena can u explain me those chess questions...
really dont understand these ones.
Consider this....
Total numbers 50-500 -> 451.
_ _ numbers that do no contain 7 4.9 = 36.
similarly _ _ _ numbers that do not contain 7 4.9.9= 324.
Therefore, numbers that contain 7 = 451-360 = 91.
Enlighten If I am wrong.
_________________
Hi Gaurav, I got the same answers. Lets wait for someone to differ. Here are some more....
6) In how many ways can we choose a black and a white square on a chess board?
(a) 32.32
(b) Other
7) In how many ways can we choose a black and a white square on a chess board if the two squares do not belong to the same row or column?
(a) 8.25
(b) Other
6) a.32.32
in chess board there are 64 squares -32 white and 32 black therefore number of ways of selecting one white is 32 and number of selecting one black sq is 32.
=32*32
7)(b)32*24
select any one white or black in 32 ways now in that row & col there will be 8 sqs of the other color...
see for ex select white square in 32 ways ok..
now in that row u'll have 4 black sqs
and in that col u'll have 4 white sqs
so from the 32 black sqs remove 8 =32-8=24
therefore 32*24
-meena
Consider this....
Total numbers 50-500 -> 451.
_ _ numbers that do no contain 7 4.9 = 36.
similarly _ _ _ numbers that do not contain 7 4.9.9= 324.
Therefore, numbers that contain 7 = 451-360 = 91.
Enlighten If I am wrong.
hey as ritika said its only 90 i think i found out ur mistake...
for _ _ _ u have not considered 500
4.9.9 are numbers only from 100 to 499....wat abt 500
4.9.9- in that 4 indicates only 1 to 4 wat abt 500?
so if u include 500 as a number then no.s (without 7)=361
so 451-361=90
see if this explanation is ok...
meena
Hey meena thanx
The explanation really helps...
A=={0,1,4,5,7,8}
Unless stated otherwise number are formed without repetition of digits. In the following question'n' refers to the number of digits in the number being formed.
Q- With repetition of digits allowed, how many numbers divisible by 4 can be formed when n=4?
A=={0,1,4,5,7,8}
Unless stated otherwise number are formed without repetition of digits. In the following question'n' refers to the number of digits in the number being formed.
Q- With repetition of digits allowed, how many numbers divisible by 4 can be formed when n=4?
just have to ascertain the 2 dig nos tht u can form with the given no:s st the no is divisible by 4..
here goes
00,04,08,40,44,48,80,84,88
no...for the remaining 2 places ...
1st place can be any of 5...(Leaving out 0...cos then it wudnt be a 4 digit no..
2nd place can be any of 6
..so totally ...5*6*9=270..
andy.
Hi All
I am a new entrant to PG. I have gone through the posts. Here are some intresting PC questions. Please give your answers and working for common benefit. Thank You.
1) How many ways can a committee of k people be chosen from 10 people if k can be 1,2,3...or 10.
(a) C(10,1)+C(10,2)+......+C(10,10)
(b) 10+10.9+10.9.8+ ..... + 10.9.8....3.2.1
(c) Other
2) A mother distributes 5 different apples among 8 children.
(i) How many ways can this be done if each child recieves at most one apple?
(a) 8.7.6 (5!)
(b) Other
(ii) How many ways can this be done if there is no restriction on the number of apples a child can recieve?
(a) 5^8
(b) Other
3) A compartment has 10 seats - 5 seats facing the engine and the other 5 facing away from the engine. Of the 10 passengers, 4 prefer to face the engine, 3 prefer away and 3 have no preference. In how many ways can the passengers be arranged on their preference?
(a) 3
(b) Other
4) In how many ways can 4 cards of different suits be selected from a deck of cards?
(a) 13^4
(b) Other
5) In (4) in how many ways can the 4 caqrds be selected where diamond selection has same value as the clubs selection and spade that of heart?
(a) 13^2
(b) Other
More to follow ....
1)(a) C(10,1)+C(10,2)+.....
2)(i)other.
the 1st apple can be given to any of the 8 children.the 2nd apple can be given to any of the remaining 7.the 3rd to any of the 6 and so on.
so total no. of ways is 8*7*6*5*4
(ii)5^8
3)other.
first arrange the 4 persons to sit facing the engine.this can be done in P(5,4) ways.now arrange the 3 persons facing away from the engine.this can be done in P(5,3) ways.now arrange the remaining 3 in 3! ways.
total no. of ways=P(5,4)*P(5,3)*3!
4)(a) 13^4
5)(a) 13^2
(ii) How many ways can this be done if there is no restriction on the number of apples a child can recieve?
(a) 5^8
(b) Other
(ii)5^8
(ii)5^8 ???????????????? y 5^8? and not 8^5?
see first apple can be given to any of the 8 chidren
2 to any of the 8
3rd to any of 8
4rth to any of 8
5th to any of 8
so 8*8*8*8*8
8^5
??????????????????????????? pelase xplain
sorry my mistake.it is 8^5 and not 5^8.
thankx
Consider this....
Total numbers 50-500 -> 451.
_ _ numbers that do no contain 7 4.9 = 36.
similarly _ _ _ numbers that do not contain 7 4.9.9= 324.
Therefore, numbers that contain 7 = 451-360 = 91.
Enlighten If I am wrong.
hey as ritika said its only 90 i think i found out ur mistake...
for _ _ _ u have not considered 500
4.9.9 are numbers only from 100 to 499....wat abt 500
4.9.9- in that 4 indicates only 1 to 4 wat abt 500?
so if u include 500 as a number then no.s (without 7)=361
so 451-361=90
see if this explanation is ok...
meena
Point taken.
Thank You Meena.
(ii) How many ways can this be done if there is no restriction on the number of apples a child can recieve?
(a) 5^8
(b) Other
(ii)5^8
(ii)5^8 ???????????????? y 5^8? and not 8^5?
see first apple can be given to any of the 8 chidren
2 to any of the 8
3rd to any of 8
4rth to any of 8
5th to any of 8
so 8*8*8*8*8
8^5
??????????????????????????? pelase xplain
Meena...this is a bit confusing to me.
In the problem it is stated that 'there is no restriction on the number of apples a child can recieve'.... which means that
child 1 can recieve 1,2,3,4 or 5 apples
child 2 5 ways ....and so on....
therefore shouldnt it be 5.5.5.5.5.5.5.5 = 5^8
kindly clarify.
The word 'INCONSISTENCY' is jumbles up in such a manner that there is no continuous formation of 'CONSISTENCY' in the so formed words. In how many ways can this be done?
hey ppl solve this with explanation..
The word 'INCONSISTENCY' is jumbles up in such a manner that there is no continuous formation of 'CONSISTENCY' in the so formed words. In how many ways can this be done?
hey ppl solve this with explanation..
first we shud find the no. of ways in which the sequence CONSISTENCY is present.
assume CONSISTENCY to be one unit(letter,since we want to have the sequence CONSISTENCY present)
so the no. of ways in which this(CONSISTENCY being present) is 3!=6
total no. of ways in which the letters of the word INCONSISTENCY can be arranged is 13!/(3!*2!*2!*2!)
the number of ways in which CONSISTENCY is not present in the arrangements is 13!/(3!*2!*2!*2!)-6
bye..
total no. of ways in which the letters of the word INCONSISTENCY can be arranged is 13!/(3!*2!*2!*2!)
hey pendyal can u elaborate on ths...
the total no. of ways of arranging n things in which m things are one type,p things of another type,q things of another type is
n!/(m!*p!*q!)
bye..
total no. of ways in which the letters of the word INCONSISTENCY can be arranged is 13!/(3!*2!*2!*2!)
hey pendyal can u elaborate on ths...
In INCONSISTENCY, there are 13 letters with
2 - I's
3 - N's
2 - C's
2 - S's
13! includes all possible permutations,
But permunations where 2 I's change their position does not give a new permutation, hence to have unique solutions we divide by 2!
Similarly by 3!-N, 2!-C and 2!-S
There for the solution 13!/(3!*2!*2!*2!)
HTH
Gaurav.
hey gaurav thanx for the explainatation
More questions from P&C;
1) How many 9 letter palindromes are possible from English Alphabets?
2) How many integers are there between 1 and 10^4 that contain exactly one 8 and one 9?
Cheers
More questions from P&C;
1) How many 9 letter palindromes are possible from English Alphabets?
2) How many integers are there between 1 and 10^4 that contain exactly one 8 and one 9?
Cheers
For the First Q
(I assume that ABCDEDCBA is a 9 letter palindrome)
SO first we need 5 letter to form a nine letter palindrome.. = 26C5
And u need to arrange the first 5 letters, others get fixed accordingly.. this can be done in 5! ways
So my answer is (26C5)*5!
Please reply if i misunderstood the question.
Haven't solved the second one yet...
Cheers
Gaurav.