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
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NA
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10
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5
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15
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72 / 100
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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!
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5
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5
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10
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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!
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12
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13
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25
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74 / 100
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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
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2
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6
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8
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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
=> 14C1*22C2+14C2*22C1+22C3 = 6776
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5
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9
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What is the number of ways of expressing 3600 as a product of three ordered positive integers (abc, bca etc. are counted as distinct).
For example, the number 12 can be expressed as a product of three ordered positive integers in 18 different ways.
3600 = 2^4 * 3^2 * 5^2
Let abc = 2^4 * 3^2 * 5^2
We have to distribute four 2's to three numbers a, b, c in (4+3-1)C(3-1)=6C2 = 15 ways.
Now two 3's has to be distributed to three numbers in (2+3-1)C(3-1)=4C2 = 6 ways
Now two 5's has to be distributed to three numbers in (2+3-1)C(3-1)=4C2 = 6 ways
Total ways = 15 * 6 * 6 = 540
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5
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How many 6 digit even numbers can be formed from digits 1, 2, 3, 4, 5, 6, and 7 so that the digit should not repeat and the second last digit is even?
Total no.of words COLLEGE can be arranged in 7!/2!*2!=1260
Vowels OEE come together = CLLG(OEE) = 5!/2!*3!/2! = 60*3 = 180
Vowels do not come together = (1260-180) = 1080
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4
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6
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In how many ways can the letter form HARDWARE do not vowels come together.
Total no of words make from HARDWARE = 8!/2!*2! = 10080
Vowels come together = 6!/2!*3!/2! = 1080
vowels nt cmg together = 8!/2!*2!-6!/2!*3!/2! = 9000
Workspace
NA
SHSTTON
0
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5
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5
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1 M:18 S
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80 / 100
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A boy has Rs 2. He wins or loses Re 1 at a time.If he wins he gets Re 1 and if he loses the game he loses Re 1. He can loose only 5 times. He is out ofthe game if he earns Rs 5. Find the number of ways in which this is possible?
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