Answer is 35
Hi,
My query is :
10 persons seated in a row find d ways in which 4 persons can be selected such dat no 2 of them are sitting nex to each oder?
Thanks
See in these type of questions first start with those which dont have restriction on them, so first place six persons like
*1*2*3*4*5*6* now remaining 4 persons can sit at any place among these 7 *.
so total ways =7c4=35
5 letter word to be constructed from MATHEMATICS,in how many words will E occur?
In how many ways can 5 letters be selected from the word INSTITUTE ?
Ans : 41
In how many ways can 5 letters be selected from the word INSTITUTE ?
Ans : 41
Yes 41 is the correct answer
consider this
I N S T I T U T E---> (I,I);(T,T,T);(N,S,U,E)
Case 1
4 are selected from NSUE
C(4,4)*2=2
Case 2
3 are selected from NSUE
Then there will be three sub cases
i) II
ii) IT
iii) TT
so C(4,3)*3=12
Case 3
2 are selected from NSUE
Then again there are 3 sub cases
i) TTT
ii)TTI
iii)IIT
Hence C(4,2)*3=18
Case 4
1 is selected from NSUE
then there are 2 cases (IITT,TTTI)
therefore C(4,1)*2=8
Case 4
0 from NSUE are selected
There is only one case IITTT
now all the above
2+12+18+8+1=41
nikita160789 Says5 letter word to be constructed from MATHEMATICS,in how many words will E occur?
is the answer 83
snipped!!!!!!
sumit99 SaysSince here we have to select 5 letters out of total 9 letters so 9c5 ways but 2 "i", 3 "t" and rest three are of 1 kind each, so divide it by 2!*3!*1*1*1 and u get 21 ways.
but the answer is 41
nikita160789 Says5 letter word to be constructed from MATHEMATICS,in how many words will E occur?
jkaustubh Saysis the answer 83
M A T H E I C S
M A T
total 11 letters : out of which 3 pairs
'E' has to be in the word
let us denote 'P-P' for a pair, and 'U' for unique
Cases:
E P-P P-P
selecting 2 pairs out of 3 pairs = 3C2 = 3 ways
arranging them in 5 letter words = 5!/(2!*2!) = 30
total ways = 90
E P-P U U
selecting 1 pair out of 3 = 3C1 = 3
selecting 2 unique out of the remaining 6 unique letters = 6C2 = 15
arranging them in 5 letter words = 5!/2! = 60
total ways = 3*15*60 = 2700
E U U U U
selecting 4 unique out of 7 = 7C4 = 35
arranging them = 5!
total ways = 35*120 = 4200
Total words having E = 90 + 2700 + 4200 = 6990
kaustubh: buddy you had considered the selection ... but you need to consider the arrangements as well
nikita160789 Says5 letter word to be constructed from MATHEMATICS,in how many words will E occur?
M M A A T T
H E I C S
when all are different = E _ _ _ _ = 7C4*5! = 120*35 = 4200
when two of them are identical = E _ _ _ _ = 3C1*6C2*5!/2! = 2700
when four of them are identical = E _ _ _ _ = 3C2*5!/2!*2! = 90
total = 4200+2700+90 = 6990
Some more queries :
1) In how many ways can a cricketor score 200 runs with fours and sixes only ?
Ans : 17 ways
2) In an exam maximum marks for each of 3 papers is 50 each. Max. marks for the 4th paper is 100. Find the no. of ways with which a student can score 60% in aggregate.
Ans : 110551
3) How many 6 digit nos. contain exactly 4 different digits ?
Ans : 294840
Some more queries :
1) In how many ways can a cricketor score 200 runs with fours and sixes only ?
Ans : 17 ways
2) In an exam maximum marks for each of 3 papers is 50 each. Max. marks for the 4th paper is 100. Find the no. of ways with which a student can score 60% in aggregate.
Ans : 110551
3) How many 6 digit nos. contain exactly 4 different digits ?
Ans : 294840
(1) Say he hits (x, y) number of (6s, 4s)
=> 6x + 4y = 200
=> 3x + 2y = 100
Now, 3x has to be even as 3x = (100 - 2y)
=> x has to be even
x can be from 0 till 32
=> 17 possible even values of x
(2) He needs to score 150 marks
=> He can lose at most 100 marks
Say, he loses (a, b, c, d) marks in 4 subjects
=> a+b+c+d = 100
=> 103C3 ways
But, a, b and c can be at most 50. We need to remove these cases.
Say, a = a'+51 where a' >= 0
=> a'+51 + b+c+d = 100
=> a'+b+c+d = 49
=> 52C3 ways
But even b or c can be more than 50
So, in all we need to remove 3*(52C3) cases
=> Total ways = 103C3 - 3*(52C3) = 110551
Note: If you are not aware of formula; please refer to this post:
http://www.pagalguy.com/forum/quantitative-questions-and-answers/72542-official-quant-thread-cat-2011-a-816.html#post2985951
(3) When 0 is not in the 4 digits; we can pick 4 digits in 9C4 ways.
Then, we can have digits as: (3, 1, 1, 1) or (2, 2, 1, 1)
(3, 1, 1, 1) => Repeating digit can be selected in 4 ways and then all can be arranged in 6!/3! ways
(2, 2, 1, 1) => The two digits can be chosen in 4C2 ways and then all can be arranged in 6!/(2!*2!) ways
=> Total cases = 9C4 * [6!/3! + 6!/(2!*2!)]
When 0 is one of the digits; we can pick other 3 digits in 9C3 ways.
Then, we can have digits as: (3, 1, 1, 1) or (2, 2, 1, 1)
(3, 1, 1, 1) => Repeating digit can be selected in 4 ways and then all can be arranged in 6!/3! ways
(2, 2, 1, 1) => The two digits can be chosen in 4C2 ways and then all can be arranged in 6!/(2!*2!) ways
Now, here we need to remove cases when 0 is at first place.
Such cases will be 1/4th as each of the 4 digits will occur at first place equal number of times.
=> Total cases = 9C3 * (3/4) * [6!/3! + 6!/(2!*2!)]
In all we have:
[ 9C4 * [6!/3! + 6!/(2!*2!)] ] + [ 9C3 * (3/4) * [6!/3! + 6!/(2!*2!)] ] = 294840 ways
There are (2n+1) numbers out of which 3numbers are to be chosen at random.What is the probability that the number choosen is in A.P.
6 white and 6 black balls of the same size are distributed among 10 urns so that there is at least one ball in each urn. what is the number of different distributions of the ball?
there are 3 consecutive odd integers such that product is prime no. find the largest of the prime no.
a.1
b.3
c.5
d.7::banghead:
there are 3 consecutive odd integers such that product is prime no. find the largest of the prime no.
a.1
b.3
c.5
d.7::banghead:
as there are 3 consecutive odd numbers so we need to have two 1's
so , numbers = -3 , -1 , 1
product = 3
largest = 1
there are 3 consecutive odd integers such that product is prime no. find the largest of the prime no.
a.1
b.3
c.5
d.7::banghead:
My take is 1 with prime number as 3
Only such triplet can be (-3, -1, 1)
In all other cases, product of 3 consecutive odd numbers cannot be a prime number.
In how many ways can you choose 6 cards out of a pack of 52 cards and have all suits present?????
:cheers:
In how many ways can you choose 6 cards out of a pack of 52 cards and have all suits present?????
:cheers:
total ways of selecting 6 cards is 52C6
now we will select 6cards of same suit...
13C6 *4
so cards having all suit present will be
52C6 - 4*13C6
ps:i have started to learn pnc...was always afraid of this topic...so never took it seriously...so plz correct me where ever i get wrong
thanx frnds
in how many ways can you choose 6 cards out of a pack of 52 cards and have all suits present?????
:cheers:
52c1*39c1*26c1*13c1*48c2