Learning these TS Inter 2nd Year Maths 2A Formulas Chapter 5 Permutations and Combinations will help students to solve mathematical problems quickly.
TS Inter 2nd Year Maths 2A Permutations and Combinations Formulas
→ Fundamental principle: If a work W1, can be performed in m different ways and another work W2 can be performed in n different ways, then the two works simultaneously can be performed in mn different ways.
→ If n is a positive integer, then n! = n {(n – 1)!} and 1! = 1.
→ We define 0! 1.
→ The number of permutations of n dissimilar things taken ‘r’ at a time is denoted by
nPr and npr = \(\frac{n !}{(n-r) !}\) for 0 ≤ r ≤ n
→ If n, r are positive integers and r ≤ n, then
- nPr = n. (n-1)P(r-1) (if r > 1)
- nPr = n.(n – 1)(n-2)P(r-2).(if r ≥ 2).
→ The number of permutations of n dissimilar things taken V at a time
- containing a particular thing is r.(n-1)P(r-1)
- not containing a particular thing is (n-1)Pr
- containing a particular thing in a particular place is (n-1)P(r-1)
→ If n, r are positive integers and r < n, then nPr = (n-1)Pr + r.(n-1)P(r-1)
→ The sum of the r-digit numbers that can be formed using the given n’ distinct non-zero digits
(r ≤ n ≤ 9) is (n-1)P(r-1) × (sum of all n digits) × (111…………….1)(rtimes)
→ In the above, if ‘0’ is one among the given ‘n’ digits, then the sum is
(n-1)P(r-1) × (sum of the digits) × (111……………1)(rtimes)
(n-1)P(r-1) × (sum of the digits) × (111………………1)(r-1)times)
→ The number of permutations of n dissimilar things taken r’ at a time when repetitions are allowed (i.e., each thing can be used any number of times) is nr.
→ The number of circular permutations of n dissimilar things is (n – 1) !.
→ In the case of hanging type circular permutations like garlands of flowers, chains of beads etc., the number of circular permutations of n things is \(\frac{1}{2}\) [(n – 1) !].
→ If in the given n things, p like things are of one kind, q alike things are of the second kind, r alike things are of the third kind and the rest are dissimilar, then the number of permutations (of these n things) is \(\frac{\mathrm{n} !}{(\mathrm{p} !)(\mathrm{q} !)(\mathrm{r} !)}\)
→ The number of combinations of n things taken ‘r’ at a time is denoted by nCr and nCr = \(\frac{n !}{(n-r) ! r !}\) for 0 ≤ r ≤ n
→ If n, r are integers and 0 ≤ r ≤ n, then nCr = nC(n-r)
→ nC0 = nCn; nC1 = nC(n-1)
→ The number of ways of dividing ‘m + n’ things (m ≠ n) into two groups containing m, n things is \(\frac{(m+n+p) !}{(m !)(n !)(p !)}\)
→ The number of ways if distributing mn things equally to m persons is \(\frac{(\mathrm{mn}) !}{(\mathrm{n} !)^{\mathrm{m}}}\)
→ If p alike things are of one kind, q alike things are of the second kind and r alike things are of the third kind, then the number of ways of selecting one or more things out of them is (p + 1) (q + 1) (r + 1) – 1.
→ If m is a positive integer and m = P1α1, P2α2 …………. Pkαk where p1, p2, ………….. , pk are distinct primes and α1, α2, …………….. αk are positive integers then the number of divisors of m is (α1 + 1) (α2 + 1) ………. (αk + 1) (This includes 1 and m).