mn into n groups
= (mn)! / (m! m! ....n times * n!(n groups are similar))
= (mn)! / (m!)^n * n!
Workspace
NA
SHSTTON
26
Solv. Corr.
16
Solv. In. Corr.
42
Attempted
0 M:0 S
Avg. Time
52 / 85
Choose the correct option.
There are 20 persons sitting in a circle. In that there are 18 men and 2 sisters. How many arrangements are possible in which the two sisters are always separated by a man?
Let the first sister name is A. Now she can sit any where in the 20 places (Symmetrical). Now her sister B can sit to her left or right in 2 ways. Now the remaining 18 persons can be sit in 18 places in 18! ways. Total = 18! * 2
Workspace
NA
SHSTTON
22
Solv. Corr.
23
Solv. In. Corr.
45
Attempted
0 M:0 S
Avg. Time
53 / 85
Choose the correct option.
In how many ways a team of 11 must be selected a team 5 men and 11 women such that the team must comprise of not more than 3 men.
5 men and 11 women and not more than 3 men
1st way= 1men and 10 women= 5C1 * 11C10
2nd way= 2 men and 9 women=5C2 *11C9
3rd way= 3 men and 8 women= 5C3* 11C8
4th way= 0 men and 11 women=5C0* 11C11
Adding all we get= 2256.
Workspace
NA
SHSTTON
8
Solv. Corr.
12
Solv. In. Corr.
20
Attempted
0 M:0 S
Avg. Time
54 / 85
Choose the correct option.
There are ten points marked on a straight line, and another 11 points marked on a line parallel to this one. How many triangles could be formed using these points ?
A manufacturer of chocolates makes 6 different flavors of chocolates. The chocolates are sold in boxes of 10.
How many "different" boxes of chocolates can be made?
If n similar articles are to be distributed to r persons, x1+x2+x3......xr=n each person is eligible to take any number of articles then the total ways are n+r?1Cr?1
In this case x1+x2+x3......x6=10
In such a case the formula for non negative integral solutions is n+r?1Cr?1 Here n = 6 and r = 10. So total ways are 10+6?1C6?1 = 3003
Workspace
NA
SHSTTON
3
Solv. Corr.
12
Solv. In. Corr.
15
Attempted
0 M:0 S
Avg. Time
56 / 85
Choose the correct option.
Seven different objects must be divided among three persons. In how many ways this can be done if at least one of them gets exactly one object.
Division of m+n+p objects into three groups is given by (m+n+p)!m!*n!*p! But 7 = 1 + 3 + 3 or 1 + 2 + 4 or 1 + 1 + 5
So The number of ways are (7)!1!*3!*3!*12! + (7)!1!*2!*4! + (7)!1!*1!*5!*12! = 70 + 105 + 21 = 196
Workspace
NA
SHSTTON
12
Solv. Corr.
13
Solv. In. Corr.
25
Attempted
0 M:0 S
Avg. Time
57 / 85
Choose the correct option.
The crew of a rowing team of 8 members is to be chosen from 12 men (M1, M2,...., M12) and 8 women (W1, W2,..., W8), such that there are two rows, each row occupying one the two sides of the boat and that each side must have 4 members including at least one women. Further it is also known W1 and M7 must be selected for one of its sides while M2, M3 and M10 must be selected for other side. What is the number of ways in which rowing team can be arranged.
We need two person for one side and 1 women for the another side. We select that women in 7 ways. Now that second side people can sit in 7*4! ways.
Now for the first side we need two people from the remaining 14. So this can be done in 14C2 ways and this side people can sit in 4C2*4! ways.
Again the first group may take any of the two sides. So total ways are 2*7*4!*14C2*4!
Workspace
NA
SHSTTON
10
Solv. Corr.
13
Solv. In. Corr.
23
Attempted
0 M:0 S
Avg. Time
58 / 85
Choose the correct option.
Mr and Mrs smith had invited 9 of their friend and their spouses for party at wiki beachresort.the stand for group photograph if mr smith never stand next to mrs smith then how many way group arrange in row.
Mr and Mrs smith never be together so out of 20 (with both them) 19 can b any place so total = 19!
20 because their invitees are with their spouses.
Also 1 of them can stand with other so 18 places
So total is 18*19!
Workspace
NA
SHSTTON
15
Solv. Corr.
20
Solv. In. Corr.
35
Attempted
0 M:0 S
Avg. Time
59 / 85
Choose the correct option.
The letters in word ADOPTS are permuted in all possible ways and arranged in the alphabetical order. Find the word at position 42 in the permuted alphabetical order.
No. of words starting with AD=4!=24, AO=4!
Thus the word has to start with AO because the last word with AO will be the 48th word.
Now no. of words formed with AOD=3!=6, AOP=3!=6, AOS=3!=6 Thus the last word formed with AOS is=4!+3!+3!+3!=24+6+6+6=42
Thus the last word formed with AOS is AOSTPD
Workspace
NA
SHSTTON
6
Solv. Corr.
11
Solv. In. Corr.
17
Attempted
0 M:0 S
Avg. Time
60 / 85
Choose the correct option.
There is a set of 36 distinct points on a plane with the following characteristics:
* There is a subset A consisting of fourteen collinear points.
* Any subset of three or more collinear points from the 36 are a subset of A.
How many distinct triangles with positive area can be formed with each of its vertices being one of the 36 points? (Two triangles are said to be distinct if at least one of the vertices is different)
The given data indicates that 14 points are collinear and remaining 22 points are non collinear.
A triangle can be formed by taking 1 points from 14 and 2 points from 22
OR
2 points from 14 and 1 points from 22
OR
3 points from 22
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