Few days back there was a Q on choosing 4 persons in four couple in such a way that NO couple is in the group and then there were Queries WHY there is certain restrictions while finding ways. Also there was a PM on grouping / combinations of dissimilar things in equal groups.I always ask students interacting with me to know HOW of a thing/process, if TIME permits. Knowing HOW can help you solve otherwise complicated Qs and SAVE time.Look at the two Q below-
1) what are the ways to distribute 8 people in group of 5 and 3? Ans 8C5 or 8C3..or say 8 people in groups of 4, 3 and 1?
Ans 8C4*4C3*1C12) what are the ways to distribute 4 people in group of 2 each? ans 4C2= 4!/2!2!=6 IS IT SO?SAY 4 people are ABCD..
groups =>
1. AB and CD
2. AC and BD
3. AD and BC
WHERE have the REMAINING 3 GONE?ANS- the other 3 became repetitions of the above threeso ans is 4C2/2!
Similarily if you were dividing 9 in three groups of 3 each, the answer will not be 9C3*6C3 but 9C3*6C3/3!..
since when we made one group of ABC, we simultaneously made DEF and GHI..Formula-
If you divide 'xa' items in 'x' groups of 'a' items each, the number of ways= \(\frac{(xa)!}{x!(a!)^x}\)lets differentiate between similar and dissimilar things-
No of balls = 4 and number of boxes=3four cases 1) All balls and All boxes are similar
2) Balls are similar BUT boxes are NOT.
3) Ball are dissimilar and Boxes are similar
4) All balls and All boxes are dissimilarLets find the solution for each case-
1) All balls and All boxes are similara) All four balls in one box-----------------
Ways = 0,0,4 = 1 wayb) 3 in one and 1 in other------------------
ways= 0,1,3 = 1 wayc) 2 in one and 2 in other------------------
ways = 0,2,2 = 1 wayd) 2 in one, 1 in second and 1 in third box--
ways= 1,1,2= 1 wayTOTAL = 4 ways2) Balls are similar BUT boxes are NOT.Now all groupings will be similar to above but the boxes are dissimilar, so lets find waysa) All four balls in one box--
Ways = 0,0,4 = 1 way ; 4 can be any of the box ------------------------------SO TOTAL = 1*3= 3ways
b) 3 in one and 1 in other--
ways= 0,1,3 = 1 way ; the three boxes can be choosen in 3! ways------------SO TOTAL = 1*3!=6 ways
c) 2 in one and 2 in other--
ways = 0,2,2= 1 way ; boxes can be choosen in 3!/2!=3 ways----------------SO TOTAL = 1*3=3 ways
d) 2 in one, 1 in second and 1 in third box--
ways= 1,1,2= 1 way ; boxes can be choosen in 3!/2!=3 ways SO TOTAL = 1*3=3 ways
TOTAL = 3+6+3+3=15 ways3) Ball are dissimilar and Boxes are similarHere boxes are the same but the balls are dissimilara) All four balls in one box--
Ways = 0,0,4 = 1 way ; since balls are in same box ----------------------------------------------------------TOTAL = 1way
b) 3 in one and 1 in other--
ways= 0,1,3 = 1 way ; balls can be choosen in 4C3 = 4 ways -----------------------------------------------TOTAL = 1*4=4 ways
c) 2 in one and 2 in other--
ways = 0,2,2= 1 way ; balls can be choosen in 4C2/2! = 3 ways (REMEMBER WHAT WE STARTED WITH) TOTAL = 1*3=3 ways
d) 2 in one, 1 in second and 1 in third box--
ways= 1,1,2= 1 way ; balls can be choosen in 4C2 = 4!/2!2!=6 ways-------------------- TOTAL = 1*6= 6 ways
TOTAL = 1+4+3+6 ways = 14 ways4) All balls and All boxes are dissimilarMANY Qs belong to this category.
After finding the ways in third, we work on the solution furthera) All four balls in one box= 1way ---
box can be any 1 of the three= 1*3 ----------------------------------------------------------------------------------------------- 3 ways
b) 3 in one and 1 in other--
balls can be choosen in 4C3 = 4 ways and Box can be choosen in 3! ways TOTAL = 1*4*6 = ---------------------------------------------24 ways
c) 2 in one and 2 in other--
balls can be choosen in \(\frac{4C2}{2!} = 3\) ways and Box can be choosen in 3! ways TOTAL = 1*3*6 = ----------18 ways
d) 2 in one, 1 in second and 1 in third box-- ;
balls can be choosen in \(4C2 = \frac{4!}{2!2!}=6\) ways and Box can be choosen in 3! ways TOTAL = 1*6*6 = 36 ways
TOTAL = 3+24+18+36 ways = 81 waysstraight formula here would be 3 can in any of boxes = 3*3*3*3=81
Whenever you look at any Q of distributing items among people or in boxes, the Q will deal with any one of the CASEs mentioned above.
LEARN to differentiate these 4 groups and you may be able to make OTHERWISE a complicated Q LOOK rather easy..Neither TIME nor SPACE permits further dwelling on the TOPIC, but you can ask any queries if you have
In 2: Identical to Distinct: I understood the method and its pretty clear. Thankyou I always struggle in these distribution questions.
we can also Use N+R-1 C R-1 formula to get the 15 ways. Are there any constraints on when to use this formula?