There are 5 blue balls, 2 red, 2 green, 1 black & 1 white. In how many ways can we select 4 out of these 11 balls ?
30
31
35
330
331
:
its 11c4 = 330...
hey!! can anybody help me wid this one...
A box contains two white balls ,three black balls and 4 red balls.In how many ways can three balls be drawn from the box if at least one black ball has to be included in the draw...pls post the approach
hey!! can anybody help me wid this one...
A box contains two white balls ,three black balls and 4 red balls.In how many ways can three balls be drawn from the box if at least one black ball has to be included in the draw...pls post the approach
Probability = 1 - number of ways when no black ball is selected
=> P = 1 - C(6, 3)/C(9, 3) = 1 - (6*4*5)/(9*8*7) = 16/21
Probability = 1 - number of ways when no black ball is selected
=> P = 1 - C(6, 3)/C(9, 3) = 1 - (6*4*5)/(9*8*7) = 16/21
I did not ask for the probability ..i wanted the approach for calculating the number of selections possible... can u pls show me how u are calculating the number of selections
hey!! can anybody help me wid this one...
A box contains two white balls ,three black balls and 4 red balls.In how many ways can three balls be drawn from the box if at least one black ball has to be included in the draw...pls post the approach
Nirvana_MBA SaysI did not ask for the probability ..i wanted the approach for calculating the number of selections possible... can u pls show me how u are calculating the number of selections
As chilly bhai has posted it easier to solve this problem taking the reverse approach...
number of ways when no black ball is selected - 6c3
here
6=>2white ball plus 4 red balls(black ball is excluded)
3=>ur selecting 3 out of 6 balls
total no of ways u can select 3 balls is - 9c3
9=>2white ball plus 4 red balls plus 3 black balls
3=>ur selecting 3 out of 9 balls
so 9c3-6c3
One more interesting question on probability:
Out of 21 tickets marked with numbers from 1 to 21 ,three are drawn at random, find the probability that the three numbers on them are in A.P
ans:- 10/133
Math Forum - Ask Dr. Math Archives: High School Permutations/Combinations
Excellent link courtesy caartey 😃 ...
Excellent link courtesy caartey 😃 ...
please let me know the solution how did u get 10/133
please write the complete method.
hello friends...!!
i recently joined the forum...nice to be here...:)
i (new to PnC) am having confusion between nCr rule and mnp rules...Ques. In how many ways can we select a 4 digit number from 1,2,3,4,5,6 without repetition?
the solution provided is 6*5*4*3...my ques is why can't we calculate as 6C4...dat is what nCr rule says(no. of ways of selecting r things from n distinct things).....
Can sum1 provide a satisfactory explanation.....
thanks in advance...:)
hello friends...!!
i recently joined the forum...nice to be here...:)
i (new to PnC) am having confusion between nCr rule and mnp rules...Ques. In how many ways can we select a 4 digit number from 1,2,3,4,5,6 without repetition?
the solution provided is 6*5*4*3...my ques is why can't we calculate as 6C4...dat is what nCr rule says(no. of ways of selecting r things from n distinct things).....
Can sum1 provide a satisfactory explanation.....
thanks in advance...:)
Hi Aditya,
Whenever you say nCr, it means that the order of arrangement is not important. The number of cases given by the formula take into account that,
1234, 1243, 2431 and all other numbers having the digits as (1,2,3,4) to be the same. So, you will get lesser number of cases.
But, if the order is important as it is in this question, you have to use nPr. That way, 1234, 1243, 2431 and other numbers having digits (1,2,3,4) will be considered as different cases.
Hope that helps.
Hi Aditya,
Whenever you say nCr, it means that the order of arrangement is not important. The number of cases given by the formula take into account that,
1234, 1243, 2431 and all other numbers having the digits as (1,2,3,4) to be the same. So, you will get lesser number of cases.
But, if the order is important as it is in this question, you have to use nPr. That way, 1234, 1243, 2431 and other numbers having digits (1,2,3,4) will be considered as different cases.
Hope that helps.
Thanx for the reply shashank,
Got this point...further, do u mean to say that mnp and nPr rule works same for these type of ques...?
1. There are seven pairs of black shoes and five pairs of white shoes. They all are put into a box and shoes are drawn ane at a time. To ensure that at least one pair of black shoes are taken out, what is the number of shoes required to be drawn out?
a) 12
b) 13
c) 7
d) 18
2. In the above question, what is the minimum number of shoes required to be drawn to get at least 1 pair of correct shoes (either white or black) ?
a) 12
b) 7
c) 13
d) 18
1. There are seven pairs of black shoes and five pairs of white shoes. They all are put into a box and shoes are drawn ane at a time. To ensure that at least one pair of black shoes are taken out, what is the number of shoes required to be drawn out?
a) 12
b) 13
c) 7
d) 18
2. In the above question, what is the minimum number of shoes required to be drawn to get at least 1 pair of correct shoes (either white or black) ?
a) 12
b) 7
c) 13
d) 18
Here u can draw all 10 white shoes first, then 7 right/left leg black shoe... So 18th shoe must form a pair..
d) 18
2)u can draw 7right/left leg black shoes and 5 right/left leg white shoes... whichever u draw the 13th time should be a pair among the two...
b) 13
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 ....
If each child has to receive atmost one, that means its different possible arrangements of 5 different apples (a,b,c,d,e) and 3 zero's.
abcde000
i.e. (8!)/(3!)
Please let me know if there's a fault in my approach.
If each child has to receive atmost one, that means its different possible arrangements of 5 different apples (a,b,c,d,e) and 3 zero's.
abcde000
i.e. (8!)/(3!)
Please let me know if there's a fault in my approach.
I guess whenever this kind of question is asked, it is safer to assume that all apples are IDENTICAL in nature unless it is specifically mentioned.
So the number of ways wud be -- 8! / (5! * 3!) = 56.
And here they are DISTINCT in nature, your answer is a perfect one i.e. 8! / 3!
Sayonara
An easier one : -
In how many ways 7^311 could be represented as the product of three natural numbers ?
Sayonara
6 people(1,2,3,4,5,6) have to sit on a circular table. 1 and 2 can never sit together and 3 and 4 have to always sit together. find the total number of arrangements.
please post the approach.
6 people(1,2,3,4,5,6) have to sit on a circular table. 1 and 2 can never sit together and 3 and 4 have to always sit together. find the total number of arrangements.
please post the approach.
See as if 3 and 4 are one and same ....
Total number of cases when 3 and 4 sit together = 4! * 2! = 48
Total no. of cases when (3 and 4) & (1 and 2) sit together = 3! * 2! * 2! = 24
There fore required no. of cases = 48 - 24 = 24
See as if 3 and 4 are one and same ....
Total number of cases when 3 and 4 sit together = 4! * 2! = 48
Total no. of cases when (3 and 4) & (1 and 2) sit together = 3! * 2! * 2! = 24
There fore required no. of cases = 48 - 24 = 24
what will be the answer if this is not a circular table but a normal linear row?