hi, please help me with the following question
how many different 4-digit numbers are there which have the digits 1,2,3,4,5,6,7 and 8 such that the digit 1 appears exactly once.
a. 7* 8P4
b. 8P4
c. 4.7^3
d. 7^3
please so mention the approach, i am getting d as an answer
Arjun verma Saysi didnt get the second part, whay calculation of zero is don seperately, kindly explain, thanks
becoz if u donot take zero separately then it will hinder u as placing zero at first place will make the number as a 3digits number( in this case) and it is included in the permutation u made... try to think practically
hi, please help me with the following question
how many different 4-digit numbers are there which have the digits 1,2,3,4,5,6,7 and 8 such that the digit 1 appears exactly once.
a. 7* 8P4
b. 8P4
c. 4.7^3
d. 7^3
please so mention the approach, i am getting d as an answer
my take on c.4.7^3
whts the OA??
Can anybody explain this problem from CL sheet
There are 7 distinct coins which r 2 b distributed randomly into 5 distinct pots in a row. Any pot can receive any no. of coins. All coins need to b distributed.
In how many ways can these coins b distributed so that no pot is empty?
Can some1 explain using the grouping approach
it must be 15
whts the OA???
your method is wrong,
it will contain some redundant numbers
like think u have numbers 5,6,6 and u are req. to make 3 digits numbers from this divisible by 5
so acc. to ur method it will be
first place can be filled by 2 ways(6,6)
second place can be filled by 1 way(6)
so total ways 2*1*1= 2
BUT THE ONLY NUMBER possible is 665
I think u must have got my point.
rethink abt it again
according to me the answer must be 249
whts the OA??
You are correct. I have failed to consider the repetitions.
it must be 15
whts the OA???
Since the coins and pots are different, answer should be
5^7 - C(5, 1)*4^7 + C(5, 2)*3^7 - C(5, 3)*2^7 + C(5, 4)*1 = 16800
5^7 is the total ways of distributing, but these includes the cases when one or more pots are empty
C(5, 1)*4^7 are the cases when atleast one pot is empty(but they includes those cases when two or more are empty)
C(5, 2)*3^7 are the cases when atleast two pots are empty(but they includes those cases when three or more are empty)
C(5, 3)*2^7 are the cases when atleast three pots are empty(but they includes those cases when four or more are empty)
C(5, 4)*1^7 are the cases when atleast four pots are empty
In a temple there r 12 pair of shoes. A boy steals 4 shoes.whats t probablty tht atleat one Left-Right pair of matchng shoes is stolen.?
a)4818/5313
b)3856/5313
c)4686/10626
d)6058/10626
In a temple there r 12 pair of shoes. A boy steals 4 shoes.whats t probablty tht atleat one Left-Right pair of matchng shoes is stolen.?
a)4818/5313
b)3856/5313
c)4686/10626
d)6058/10626
Lets find the number of cases when none of the matching left right pair is chosen. First card can be chosen in 24 ways, for second we have 22 possibilities as pair of the first shoe can not be chosen, similarly for third we have 20 possibilities and last we have 18 possibilities. But here order is not important so we need to divide by 24.
=> No of ways = (24*22*20*18 )/24 = 7920
=> Required no of ways = C(24, 4) - 7920 = 2706
=> Probability = 2706/10626
Alternatively:-
There are two possibilities:-
i) When two pairs are chosen
C(12, 2) cases
ii) When only one pair is chosen
C(12, 1)*22*20/2 (one pair can be chosen in 12 ways and third can be chosen in 22 ways and last one in 20 ways, but order is not important so we need to divide by 2)
=> Total cases = 66 + 12*22*10 = 2706
=> Probability = 2706/10626
In a temple there r 12 pair of shoes. A boy steals 4 shoes.whats t probablty tht atleat one Left-Right pair of matchng shoes is stolen.?
a)4818/5313
b)3856/5313
c)4686/10626
d)6058/10626
Total number of ways =(24 C 4)=10626
Total number of ways in which we could not select a pair = 24*22*20*18 /24 = 7920
Total number of favourable condition = 10626-7920 = 2706
So, Probability = 2706/7920
my take on c.4.7^3
whts the OA??
yes, your answer is write, please explain it
Arjun verma Saysyes, your answer is write, please explain it :lookround:
since 1 can only appear one time so 1 cannot occur anymore
numbers left are 2,3,4,5,6,7,8 ( seven numbers 7)
fixing 1 at first place
other three places can be occupied by any of the seven numbers, any number of times so 7^3
similarly fixing 1 at second, third and fourth place will also give 7^3
hence, 4 * 7^3
hope i made myself clear...???
five boys and three girls are sitting in a row of eight seats. in how many ways can they be seated so that not all girl sit side by side.
ans: 36000
please do post your approoach
Since the coins and pots are different, answer should be
5^7 - C(5, 1)*4^7 + C(5, 2)*3^7 - C(5, 3)*2^7 + C(5, 4)*1 = 16800
5^7 is the total ways of distributing, but these includes the cases when one or more pots are empty
C(5, 1)*4^7 are the cases when atleast one pot is empty(but they includes those cases when two or more are empty)
C(5, 2)*3^7 are the cases when atleast two pots are empty(but they includes those cases when three or more are empty)
C(5, 3)*2^7 are the cases when atleast three pots are empty(but they includes those cases when four or more are empty)
C(5, 4)*1^7 are the cases when atleast four pots are empty
we can also do by taking sets
7 coins can be distributed as (1,1,1,1,3) or (1,1,1,2,2)
but not able to proceed any further. please post the approach Chill sir.
five boys and three girls are sitting in a row of eight seats. In how many ways can they be seated so that not all girl sit side by side.
Ans: 36000
please so post your approoach
8!-6!*3!=36000
2. Seven different objects must be divided among three people. In how many ways can this be done if one or two of them must get no objects?
ans: 381
3. Seven different objects must be divided among three people.In how many ways can this be done if at least one of them gets exactly one object?
ans: 1176
please do post your approach.
2. Seven different objects must be divided among three people. In how many ways can this be done if one or two of them must get no objects?
ans: 381
3. Seven different objects must be divided among three people.In how many ways can this be done if at least one of them gets exactly one object?
ans: 1176
2) 3*2^7 - 3 = 384 - 3 = 381
3*2^7 are the cases when atleast one of them haven't got anything, but few cases are repeated since here all those cases are also there when two of them haven't got anything
3 are the cases when only one of them got all objects
_______________________________________________________________
3) Three cases:-
i) (1, 1, 5) - 3*C(7, 1)*C(6, 1) = 126 ways
3 because 5 can be given to any one of the three
ii) (1, 2, 4) - 6*C(7, 1)*C(6, 2) = 630
6 because (1, 2, 4) can be permuted in 3! ways for three persons
iii) (1, 3, 3) - 3*C(7, 1)*C(6, 3) = 420
3 because 1 can be given to any one of the three
Total = 126 + 630 + 420 = 1176
Lets find the number of cases when none of the matching left right pair is chosen. First card can be chosen in 24 ways, for second we have 22 possibilities as pair of the first shoe can not be chosen, similarly for third we have 20 possibilities and last we have 18 possibilities. But here order is not important so we need to divide by 24.
=> No of ways = (24*22*20*18 )/24 = 7920
=> Required no of ways = C(24, 4) - 7920 = 2706
=> Probability = 2706/10626
Alternatively:-
There are two possibilities:-
i) When two pairs are chosen
C(12, 2) cases
ii) When only one pair is chosen
C(12, 1)*22*20/2 (one pair can be chosen in 12 ways and third can be chosen in 22 ways and last one in 20 ways, but order is not important so we need to divide by 2)
=> Total cases = 66 + 12*22*10 = 2706
=> Probability = 2706/10626
Hi chillfactor!!
Can u tell me when order is not important then why we need to divide it by 4!
Plz explain elaborately
Thanks
three groups of children 3 girls and 1 boy, 2 girls and 2 boys, and 1 girl and 2 boys repectively. one child is selected at random from each group. the probability that the three selected consist of 1 girl and 2 boys is?
ans:3/8
please do post your approach.
three groups of children 3 girls and 1 boy, 2 girls and 2 boys, and 1 girl and 2 boys repectively. one child is selected at random from each group. the probability that the three selected consist of 1 girl and 2 boys is?
ans:3/8
please do post your approach.
Probability of choosing girl from first gp = 3/4
Probability of choosing girl from second gp = 1/2
Probability of choosing girl from third gp = 1/3
Probability of choosing boy from first gp = 1/4
Probability of choosing boy from second gp = 1/2
Probability of choosing boy from third gp = 2/3
Now there can be three cases as the one girl can be chosen from any of the three groups, so
P = 3/4*1/2*2/3 + 1/2*1/4*2/3 + 1/3*1/4*1/2 = 9/24 = 3/8
Hi chillfactor!!
Can u tell me when order is not important then why we need to divide it by 4!
Plz explain elaborately
Thanks
Well, suppose in one case we have selected a first, then b, then c and then d, while in other case the order is b, c, d, a. but if we look at the final selections both are same.
So, when order is not important we have to divide by 4!, as a, b, c, d can be permuted in 4! ways
I hope it makes some sense
a fair dice is tossed six times. find the probability of getting a third six on the sixth throw
ans: (5C2*5^3)/6^6
please do mention your approach