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amit

· started a discussion

· 1 Months ago

doubt (Question: 90362.0 )

provide another solution

Question:
Harpreet has 22 identical red balls and 15 identical green balls. In how many ways he can put all ‘these balls in two identical baskets if no basket remains empty?
Options:
A) 192
B) 154
C) 183
D) 184 
E) None of these
Solution:

Red = 22

Green = 15


Let us first look at the distribution of A balls. Red balls can be placed as shown here.

Except for the first case that is when one box has 0 red balls and other has 22 red balls, and the last case when each of the boxes has 11 balls, the green balls can be placed in 16 ways.

That gives 10 × 16 = 160 ways and for the first e green balls can be placed in 15 ways.

Also note that, for the case when each box has 1’ balls, then the green balls can be placed in only 8 and not 16 ways.

∴ Total number of ways = 160 + 15 + 8 = 183


Alternate solution: 


RED= 0, 1 ......... 22 = 23 case

GREEN = 0,1.........15= 16 case

Total case = 23*16 = 368 

baskets are identical = 368/2 = 184

But no basket remains empty = 184-1 = 183


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