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5 Most important question regarding permutations and combinations.

1.Find out the number of words which can form the latter of word 'GENERAL' if repetiation is not allowed.

Solution: Given that a word'GENERAL'in which we have 6 distinct latter such that G,E,N,R,A and L

We want to form word in which 7 latter and repetiation is not allowed.

So we have 7 place such that _ _ _ _ _ _ _

Now the number of way to fill first place = 7

Number of way to fill second place = 6

Number of way to fill the third place = 5

Number of way to fill fourth place = 4

Number of way to fill fifth place = 3

Number of way to fill sixth place = 2

And number of way to fill seventh place = 1

So required number of words = 7×6×5×4×3×2×1 = 5040.

But here both 'E'are same.

Hence the total number of words = 5040/2! = 5040/2 = 2520.

2.Find out the number of words which can form the latter of word 'CALCULUS' if repetiation is not allowed.

Solution: Given that a word 'CALCULUS' in which we have 5 distinct latter such that C,A,L,U, and S.

We want to form word in which 8 latter and repetiation is not allowed.

So we have 8 place such that _ _ _ _ _ _ _ _

Now the number of way to fill first place = 8

Number of way to fill second place = 7

Number of way to fill the third place = 6

Number of way to fill fourth place = 5

Number of way to fill fifth place = 4

Number of way to fill sixth place = 3

Number of way to fill seventh place = 2

And number of way to fill eight place = 1

So required number of words = 8×7×6×5×4×3×2×1= 40320

But here both 'C' , both 'L' and both 'U' are same.

Hence the total number of words =40320/2!×2!×2! = 40320/2×2×2 = 5040.

3.Find out the number of words which can form the latter of word 'CLASS' if repetiation is not allowed.

Solution: Given that a word 'CLASS' in which we have 4 distinct latter such that C,L,A and S.

We want to form word in which 5 latter and repetiation is not allowed.

So we have 5 place such that _ _ _ _ _ 

Now the number of way to fill first place = 5

Number of way to fill second place = 4

Number of way to fill the third place = 3

Number of way to fill fourth place = 2

And Number of way to fill fifth place = 1

So required number of words = 5×4×3×2×1 = 120.

But here both 'S' are same.

Hence the total number of words = 120/2! = 120/2 = 60.

4.Find out the number of words which can form the latter of word 'TWITTER' if repetiation is not allowed.

Solution: Given that a word'TWITTER'in which we have 5 distinct latter such that T,W,I,E and R.

We want to form word in which 7 latter and repetiation is not allowed.

So we have 7 place such that _ _ _ _ _ _ _

Now the number of way to fill first place = 7

Number of way to fill second place = 6

Number of way to fill the third place = 5

Number of way to fill fourth place = 4

Number of way to fill fifth place = 3

Number of way to fill sixth place = 2

And number of way to fill seventh place = 1

So required number of words = 7×6×5×4×3×2×1 = 5040.

But here three 'T'are same.

Hence the total number of words = 5040/3! = 5040/6 = 840.

5.Find out the number of words which can form the word 'CIRCULAR' if repetiation is not allowed.

Solution: Given that a word 'CIRCULAR' in which we have 7 distinct latter such that C,I,R,U, L,Aand R.

We want to form word in which 8 latter and repetiation is not allowed.

So we have 8 place such that _ _ _ _ _ _ _ _

Now the number of way to fill first place = 8

Number of way to fill second place = 7

Number of way to fill the third place = 6

Number of way to fill fourth place = 5

Number of way to fill fifth place = 4

Number of way to fill sixth place = 3 

Number of way to fill seventh place = 2

And Number of way to fill eight place = 1

So required number number of words = 8×7×6×5×4×3×2×1 = 40320

But here both 'C' are same.

Hence the total number of words = 40320/2!= 40320/2 = 20160.

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