But, it was never asked to seat men & women in alternate chairs in the Q.
even the combination of men is not required since the word 'particular' is used and not 'any'.
So, I think the whole lot of 12 need to be considered as 12 diff individuals.....
Wat say ?
Q39.How many different nine digit numbers can be formed from the number
223355888 by rearranging its digits so that the odd digits occupy even positions?
(a) 16(b) 36 (c) 60
((4!/2!*2!*2!)*2!)*(5!/2!*3!)=60
C is correct.
a+b+c+d+e+f=10
15c5
I think u are mistaken, when it is mentioned that no variable can have zero, so the answer will be (10 +5 -6) C (5) = 9C5
please help....
In how many ways 6 men and 6 women can be seated in a row such that 2 particular men are always together and 2 particular women are never together..
My take is 9!*2!*10*2!
i am revising the previous mocks and came across a qn( AIMCAT 1206)
i dont remember the exact wording now but it was like this.
6 people have to be distributed in 3 diff cars with atleast 1 in each.
i could solve it till this point.
the no of diff ways are
4,1,1
2,2,2
3,2,1
after this i am not getting the arrangement part. somebody plz
explain .
thanks in advance...
I think the answer is 5C2
please help....
In how many ways 6 men and 6 women can be seated in a row such that 2 particular men are always together and 2 particular women are never together..
my answer is 11! *2 - 10
i am revising the previous mocks and came across a qn( AIMCAT 1206)
i dont remember the exact wording now but it was like this.
6 people have to be distributed in 3 diff cars with atleast 1 in each.
i could solve it till this point.
the no of diff ways are
4,1,1
2,2,2
3,2,1
after this i am not getting the arrangement part. somebody plz
explain .
thanks in advance...
Till now you are right !!
1st case 4,1,1: 6C4*3!/2!
2nd case 2,2,2: (6!/2!/2!/2!/3!)*3!
3rd case 3,2,1: 6C3*3C2*3!
Add all the cases you will get the ans.:cheerio:
help me with the following problem:
The number of four digit numbers strictly greater than 4321 that can be formed from the digits 0,1,2,3,4,5 allowing for repetition of digits is
a)310 b)360 c)288 d)300
help me with the following problem:
The number of four digit numbers strictly greater than 4321 that can be formed from the digits 0,1,2,3,4,5 allowing for repetition of digits is
a)310 b)360 c)288 d)300
Ans. is 288
help me with the following problem:
The number of four digit numbers strictly greater than 4321 that can be formed from the digits 0,1,2,3,4,5 allowing for repetition of digits is
a)310 b)360 c)288 d)300
My take is 310.
When first digit is 5, last 3 digits can be filled in 6 ways each => 6*6*6 = 216 numbers
When first digit is 4 and second digit is 4 or 5; last two digits can be filled in 6 ways each => 2*6*6 = 72 numbers
When first digit is 4 and second is 3; third digit can be filled with 3, 4, or 5 and last one in 6 ways => 18 numbers or thrid digit can be 2 and last digit can be 2 to 5 => 4 numbers
So, total 216 + 72 + 18 + 4 = 310 numbers
help me with the following problem:
The number of four digit numbers strictly greater than 4321 that can be formed from the digits 0,1,2,3,4,5 allowing for repetition of digits is
a)310 b)360 c)288 d)300
Total 4 digit numbers = 5*6*6*6 = 1080
Numbers starting with 1 or 2 or 3 = 3*6*6*6 = 648
Starting with 40 or 41 or 42 = 3*6*6 = 108
Starting with 430 or 431 = 2*6 = 12
Starting with 432 = 2
So, 1080 - 648 - 108 - 14 = 310
Hey Guys
Here r my problems from P&C...Got; screwed up these days in this chapter.. 😞 :- :banghead:
Ques1:- India plays two matches each with NZ & SA.In any match,probability of different outcomes for India is given below:-
OUTCOME WIN LOSS DRAW
Probability 0.5 0.45 0.05
Points 2 0 1
Outcome of all the matches are independent of each other.
Q- What is the probability of India getting at least 7points in contest? Assume SA & NZ play 2 matches.
(a)0.025 (b)0.0875
(c)0.0625 (d)0.975
Q-What is the probability of SA getting at least 4 points? Assume SA & NZ play 2 matches.
(a)0.2025 (b)0.0625
(c)0.0425 (d)Cant be determined
-------------------------------------------------------------------
Ques2:- m men & n women are to be seated in a row so that no two women seats together. If m>n,then find the no. of ways in wich they can b seated.
(a)m!*m+1Pn
(b)m!*m+1Cn
(c)m!*mPn
(d)m!*mCn
-------------------------------------------------------------------
P.S: Please provide answers with Explaination TIA
help me with the following problem:
The number of four digit numbers strictly greater than 4321 that can be formed from the digits 0,1,2,3,4,5 allowing for repetition of digits is
a)310 b)360 c)288 d)300
My Take is 310.
That is:- For 43_ _ = 22No.s
For44 _ _ =36No.s
For45_ _ =36No.s
For 5_ _ _ =216No.s
Therefore Total =310 No.s
i know my doubt is irrelevant to dis thread since its abt calculatns...
i find it time consuming wen i come acrosss a square-root or cube root in a calculation..suppose x= cube root of say 99,,how can i increase my speed...i mean can anyone help me wid a trick short-cut method o smthing similar??
thanx in advance
Ten books are arranged in a row on a shelf. In how many ways can 3 books be selected simultaneously from the shelf such that no two adjacent books from the shelf are chosen?
1)36
2)48
3)42
4)56
5)84
I have a solution and answer but need the approach. Thanks.
Ten books are arranged in a row on a shelf. In how many ways can 3 books be selected simultaneously from the shelf such that no two adjacent books from the shelf are chosen?
1)36
2)48
3)42
4)56
5)84
I have a solution and answer but need the approach. Thanks.
It should be 56.
We have to pick 3 books, say, B1, B2 and B3.
So, a B1 b B2 c B3 d ... where, a, b, c and d are number of books at those places.
Now, a+b+c+d = 10-3 = 7
But, b and c cannot be 0.
So, a+b'+c'+d = 5 .... b = b'1+1 and c = c'1+1
=> 8C5 ways = 56 ways
Ten books are arranged in a row on a shelf. In how many ways can 3 books be selected simultaneously from the shelf such that no two adjacent books from the shelf are chosen?
1)36
2)48
3)42
4)56
5)84
I have a solution and answer but need the approach. Thanks.
Another approach:
Initially place the seven books..
-B1-B2-B3-B4-B5-B6-B7-
So now it creates 8 gaps where that three books can be placed..
So its 8c3=56..
help me with the following problem:
The number of four digit numbers strictly greater than 4321 that can be formed from the digits 0,1,2,3,4,5 allowing for repetition of digits is
a)310 b)360 c)288 d)300
5_ _ _ (6*6*6)
432 _ ( 4)
43_ _ ( 3*6)
4_ _ _ (2*6*6)
total = 310 ways
i accept neetu jain method. it is 310 ways.
help me with the following problem:
The number of four digit numbers strictly greater than 4321 that can be formed from the digits 0,1,2,3,4,5 allowing for repetition of digits is
a)310 b)360 c)288 d)300
4321
432_ = 4c1 = 4
433_ = 3*6c1 = 18
44_ _ = 2*6c1*6c1 = 72
5_ _ _ = 6c1*6c1*6c1= 216
=> 4+18+72+216 = 310