Three ladies have each brought a child for admission to a school. The head of the school wishes to interview six people one by one, taking care that no child is interviewed before his/her mother. In how many different ways interviewed can be arranged?
@Rshmi05
from 1 to 1000 the three digit no. with all 5 s is 555
2 or 3 digit nos with 2 5s are 9*3=27
2 or 3 digit nos with 1 5 are 9*9*3=243
Total 1*3 + 27*2 +243=300 5s will be there
Subtract 1 to take out single 5 we get 299 as the answer.
@sidroy09 said: Three ladies have each brought a child for admission to a school. The head of the school wishes to interview six people one by one, taking care that no child is interviewed before his/her mother. In how many different ways interviewed can be arranged?
In these type of questions You can directly do like this:
Total persons ---> 6
Total persons in a pair --> 2 (Mom and Child)
So Ways = 6!/(2!*2!*2!) = 90 [Its the best approach]
For in detail approach go here:
http://arshadmajeed.in/yahoo_site_admin/assets/docs/PERMUTATIONSCOMBINATIONSBPast_Year_IIT_questions.8652545.doc
@tempo said: If we place 1 ball each ( which is mandatory) in 10 urns ( this can be done in 1 way only)Now, 2 balls remain....for first amongst the 2 rem. balls, 10 choices available..10 ways and for 2nd...again 10 choices....10 waysSo total ways = 1 x 10 x 10 = 100
i got 300 as ans.
bcz placing 10 balls out of 12 into 10 urns can be done in 3 ways -
bcz placing 10 balls out of 12 into 10 urns can be done in 3 ways -
1) 6W 4B
2) 4W 6B
3) 5W 5B
therefore, 3 x 100 = 300
@Aizen said:In these type of questions You can directly do like this:Total persons ---> 6Total persons in a pair --> 2 (Mom and Child)So Ways = 6!/(2!*2!*2!) = 90 [Its the best approach]For in detail approach go here:Thanx.... But how come u derived the
6!/(2!*2!*2!) ?
@sachisurbhi Considering all possibilities : 8C4 * 5C3 = 700
Then , the cases in which Balamurali and science professor are together = 7C3 * 4C2 = 210.
So , now the answer will be , 700 - 210 = 490
Can Anyone explain beggar coin problems?
Please solve this and explain...............
In how many wayscan 12 similar chocolates be distributed among 3 persons where each person can get a maximum of 6 chocolates.
@Rshmi05 said:Please solve this and explain...............
In how many wayscan 12 similar chocolates be distributed among 3 persons where each person can get a maximum of 6 chocolates.
First distribute 6 chocolates each to all the three students. Now they have 18 chocolates in total.
Now you need to pick 6 chocolates out of these 3 groups
which is the solution of a + b + c = 6 = 8!/6!2! = 28
Now you need to pick 6 chocolates out of these 3 groups
which is the solution of a + b + c = 6 = 8!/6!2! = 28
If four squares of size 1*1 are chosen at random a chess board of size 8*8, what is the probability that they are in same diagonal?
Please anyone give a detailed and lucid explanation .
@rajatt1557
(4(7C4+6C4+5C4+4C4)+2*8C4)/64C4
Please solve this que:
66 students of a class are covered under a Health Insurance Scheme for which premium charged is Rs. 5000 per student. Considering the types of coverage in the policy, the probability that a policy will become a claim policy is estimated to be 0.2. The Insurance Company offering the cover has collected the data related to treatment expenses of various diseases covered under the scheme. Based on this, the claim amount is likely to vary linearly between Rs. 10000 to Rs. 50000.
What is the probability that this scheme will result in a loss. The company keeps 20% of the premium amount collected for meeting overhead and other expenses and only 80% is available for claim settlement.
The Insurance Company wants to do such business only if the probability of loss is less than or equal to 20%.
Is the probability of loss in this case
I think the insurance company should charge Rs. 10500 as premium to keep the probability of loss under 20 %. Originally, the probability of loss is 75%
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?
how can we solve this with gap method????