is the ans to the first ques
101C2 - 352C2 = 1275
352C2 .. What for is this part? can you explain your solution please.
"ways to choose 3 balls of all
minus ways not to choose any black ball "
i.e., 9C3-6C3 = 64 (Balls of same color are different)
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For the 1st question,
Let the shares to each son be S1, S2, S3.
S1 + S2 + S3 = 101
No. of integral solutions=100C2
Now, to remove those cases where one son gets more shares than the combined total of other two,
S1 + S2 ≤ 50
or,
S1 + S2 + x = 50
x ≥ 0
S1 = 1+ a [a≥0]
S2 = 1+ b [b≥0]
therefore,
a + b + x = 48
No . of integral solutions = 50C2
There will be three such cases.
So, no of ways to divide the shares = 100C2 - 3( 50C2)
choosing two numbers from 1,2,3 where repetitions are allowed
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