Combinatorics Question?

1)

Out of p alike things, we may choose one or more or none. Thus, there are (p+1) ways of choosing none or one or more of p things. If * denotes that none of the p things are chosen, these combinations can be expressed as {*}, {p}, {pp}, {ppp}, ... {ppp upto p things}.

Similarly, there are (q+1) ways of choosing or not choosing one or more of q alike things and so on.

For t different things, we have two choices for each of them, choosing or not choosing which means that there are 2^t ways of choosing or not choosing each of t different things.

Thus, there are (p+1)(q+1)(r+1)(s+1)(2^t) ways of choosing things out of n things which include a possibility of not choosing any. Hence, the total number of ways of choosing excluding a combination in which none is chosen is

(p+1)(q+1)(r+1)(s+1)(2^t) - 1.

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q2

2 distinguishable, 3 identical to 4 people

4^2 ways for distinguishable,

C(6,3) for identical

multiply to get ans

p distinguishable, q identical to r people

r^p *C(q+r-1,r-1)

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