How many possible cases are there such that 2 stocks are selected at r

Number of possible cases = 10c2
=10!/(8!*2!)
=10*9/2
=45
Answer C

Method 1:
No. of ways of selecting two out of 10 stocks = 10C2 = 10!/(2!*8!) = 45

Method 2:
No. of ways of making pair of one stock with each of the remaining stock = 9
Total Such possible pairs (with Repetition) = 10*9 = 90
Since every case has been counted twice therefore total Unique pairs = 10*9 / 2 = 45

Method 3:
Pairs of first stock with each of the 9 stocks can be made in 9 ways

Pairs of Second stock with each of the remaining 8 stocks can be made in 8 ways (second has already made pair
with first so we will not make that pair again

Pairs of Third stock with each of the remaining 7 stocks can be made in 7 ways (Third has already made pair with first and second so we will not make those pairs again) .... and so on...

Total Such pairs = 9+8+7+6+5+4+3+2+1 = 45

Answer: option C

How many possible cases are there such that 2 stocks are selected at random from 10 stocks?

A. 35
B. 40
C. 45
D. 50
E. 55

->10C2=10*9/2*1=45. Thus, the answer is C.

sir what is the difference between random and simultaneously. can we use 10c1x9c1 in random case? please explain

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