Fundamental Principle of Counting
Consider, Mohan has 3 pants and 2 shirts.
Let us name the three pants as P1,P2,P3 and the two shirts as S1,S2.
There are 3 ways in which a pant can be chosen(P1/P2/P3).
Similarly, a shirt can be chosen in 2 ways(S1/S2).
For every choice of pant, there are 2 choices of a shirt (P1 and S1/S2) (P2 and S1/S2) (P3 and S1/S2).
Therefore, there are 3*2=6 pairs of a pant and a shirt.
Consider, Sabnam has 2 school bags, 3 tiffin boxes and 2 water bottles.
Let us name the 2 school bags as B1,B2 and the 3 tiffin boxes as T1,T2,T3 and 2 water bottles as W1,W2.
A school bag can be chosen in 2 ways (B1/B2).
Tiffin box can be chosen in 3 ways(T1/T2/T3).
Water bottle can be chosen in 2 ways(W1/W2).
For every choice of school bag, there are 3 choices of a tiffin box (B1 and T1/T2/T3) (B2 and T1/T2/T3).
Therefore, there are 2*3=6 pairs of a school bag and a tiffin box.
For each of these 6 pairs, a water bottle can be chosen in 2 ways (B1 and T1/T2/T3 and W1/W2) (B2 and T1/T2/T3 and W1/W2)
Therefore, there are 6*2=12 different ways of choosing.
"If an event can occur in $m$ different ways, following which another event can occur in $n$ different ways, then the total number of occurrence of the events in the given order is $m*n$."
"If an event can occur in $m$ different ways, following which another event can occur in $n$ different ways, following which a third event can occur in $p$ different ways, then the total number of occurrence of the events in the given order is $m*n*p$."
Example 1: 3 persons enter a railway carriage, where there are 5 vacant seats. In how many ways can they seat themselves?
First man can sit on any of 5 vacant seats.
Second can sit on any of 4 vacant seats left.
Third can sit on any of 3 vacant seats left.
Hence, by fundamental principle of counting, the required number of ways=5*4*3=60.
Example 2: How many 4 letter code words are possible using the first 10 letters of the English alphabet if (i) no letter can be repeated? (ii) letters may be repeated?
(i) No letter can be repeated
First letter of code can be chosen in 10 ways.
Second letter of code can be chosen in 9 ways.
Third letter of code can be chosen in 8 ways.
Fourth letter of code can be chosen in 7 ways.
Therefore, the total number of 4 letter code=10*9*8*7=5040.
(ii) letters may be repeated
First letter of code can be chosen in 10 ways.
Second letter of code can be chosen in 10 ways.
Third letter of code can be chosen in 10 ways.
Fourth letter of code can be chosen in 10 ways.
Therefore, the total number of 4 letter code=10*10*10*10=10000.
Consider, Mohan has 3 pants and 2 shirts.
Let us name the three pants as P1,P2,P3 and the two shirts as S1,S2.
There are 3 ways in which a pant can be chosen(P1/P2/P3).
Similarly, a shirt can be chosen in 2 ways(S1/S2).
For every choice of pant, there are 2 choices of a shirt (P1 and S1/S2) (P2 and S1/S2) (P3 and S1/S2).
Therefore, there are 3*2=6 pairs of a pant and a shirt.
Consider, Sabnam has 2 school bags, 3 tiffin boxes and 2 water bottles.
Let us name the 2 school bags as B1,B2 and the 3 tiffin boxes as T1,T2,T3 and 2 water bottles as W1,W2.
A school bag can be chosen in 2 ways (B1/B2).
Tiffin box can be chosen in 3 ways(T1/T2/T3).
Water bottle can be chosen in 2 ways(W1/W2).
For every choice of school bag, there are 3 choices of a tiffin box (B1 and T1/T2/T3) (B2 and T1/T2/T3).
Therefore, there are 2*3=6 pairs of a school bag and a tiffin box.
For each of these 6 pairs, a water bottle can be chosen in 2 ways (B1 and T1/T2/T3 and W1/W2) (B2 and T1/T2/T3 and W1/W2)
Therefore, there are 6*2=12 different ways of choosing.
"If an event can occur in $m$ different ways, following which another event can occur in $n$ different ways, then the total number of occurrence of the events in the given order is $m*n$."
"If an event can occur in $m$ different ways, following which another event can occur in $n$ different ways, following which a third event can occur in $p$ different ways, then the total number of occurrence of the events in the given order is $m*n*p$."
Example 1: 3 persons enter a railway carriage, where there are 5 vacant seats. In how many ways can they seat themselves?
First man can sit on any of 5 vacant seats.
Second can sit on any of 4 vacant seats left.
Third can sit on any of 3 vacant seats left.
Hence, by fundamental principle of counting, the required number of ways=5*4*3=60.
Example 2: How many 4 letter code words are possible using the first 10 letters of the English alphabet if (i) no letter can be repeated? (ii) letters may be repeated?
(i) No letter can be repeated
First letter of code can be chosen in 10 ways.
Second letter of code can be chosen in 9 ways.
Third letter of code can be chosen in 8 ways.
Fourth letter of code can be chosen in 7 ways.
Therefore, the total number of 4 letter code=10*9*8*7=5040.
(ii) letters may be repeated
First letter of code can be chosen in 10 ways.
Second letter of code can be chosen in 10 ways.
Third letter of code can be chosen in 10 ways.
Fourth letter of code can be chosen in 10 ways.
Therefore, the total number of 4 letter code=10*10*10*10=10000.
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