Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is:
A.
880
B.
629
C.
630
D.
879
Answer :
879
Solution :
Number of white balls $$= 10$$
Number of green balls $$= 9$$
and Number of black balls $$= 7$$
∴ Required probability $$= (10 + 1) (9 + 1)(7 + 1) - 1 = 11 \cdot 10 \cdot 8 - 1 = 879$$
[ $$\because $$ The total number of ways of selecting one or more items from $$p$$ identical items of one kind, $$q$$ identical items of second kind; $$r$$ identical items of third kind is $$(p + 1)(q + 1 )(r + 1) - 1 ] $$
Releted MCQ Question on Algebra >> Permutation and Combination
Releted Question 1
$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$ and $$^n{C_{r + 1}} = 126,$$ then $$r$$ is:
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated are
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