# Inter 2nd Year Maths 2A Permutations and Combinations Formulas

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 5 Permutations and Combinations to solve questions creatively.

## Intermediate 2nd Year Maths 2A Permutations and Combinations Formulas

→ Fundamental principle: If a work can be performed in m different ways and another work can be performed in n different ways, then the two works simultaneously can be performed in mn different ways.

→ If n is a non-negative integer then

• 0 ! = 1
• n ! = n (n – 1) ! if n > 0

→ 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 positive integers and r ≤ n, then

• nPr = n. n – 1Pr – 1 if r ≥ 1
• nPr = n.(n – 1) n – 2Pr – 2 if r ≥ 2
• nPr = n – 1Pr + r. (n – 1)P(r – 1)

→ The number of injections that can be defined from set A into set B is n(B)pn(A) n(A) ≤ n(B)

→ The number of bijections that can be defined from set A into set B having same number of elements with A is n(A)!

→ The number of permutations of1n’ dissimilar things taken ‘r’ at a time.

• Containing a particular thing is (r) n – 1Pr – 1
• Not containing a particular thing is n – 1Pr
• Containing a particular thing in a particular place is n – 1Pr – i.

→ The number of functions that can be defined from set A into set B in [n(B)]n(A)

→ The sum of the r – digited 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 (r times)

→ 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 (r times) (n – 2)P(r – 2) × 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 any number of times is nr.

→ The number of circular permutations of n dissimilar things is (n – 1) !

→ In 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 alike 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 $$\frac{n !}{p ! q ! r !}$$ → 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 and $$\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{n-r+1}{r}$$ and nPr = r! nCr

→ If n, r are integers and 0 ≤ r ≤ n then ncr = nCr – 1

→ Let n, r, s are integers and 0 ≤ r ≤ n, 0 ≤ s ≤ n. If nCr = nCs then r = s or r + s = n.

→ The number of ways of dividing ‘m + n’ things (m ≠ n) into two groups containing m, n things is (m + n)Cm = (m + n)Cm = $$\frac{(m+n) !}{m ! n !}$$

→ The number of ways of dividing (m + n + p) things (m, n, p are distinct) into 3 groups of m, n, p things is $$\frac{(m+n+p) !}{m ! n ! p !}$$.

→ The number of ways of dividing mn things into m equal groups is $$\frac{(m n) !}{(n !)^{m} m !}$$

→ The number of ways if distributing mn things equally to m persons is $$\frac{(m n) !}{(n !)^{m}}$$

→ 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, p1α2 …… pkαk where p1, p2 …… pk are distinct primes and α1, α2, …. αk are non-negative integers, then the number of divisors of m is (α1 + 1) (α2 + 1) ……… (αk + 1). [This includes 1 and m]. → The total number of combinations of n different things taken any number of times is 2n.

→ The total number of combinations of n different things taken one or more at a timers 2n – 1.

→ The number of diagonals in a regular polygon of n sides is $$\frac{n(n-3)}{2}$$

→ Permutations are arrangements of things taken some or all at a time.

→ In a permutation, order of the things is taken into consideration.

npr represents the number of permutations (without repetitions) of n dissimilar things taken r at a time.

→ Fundamental Principle: If an event is done in ‘m’ ways and another event is done in ‘n’ ways, then the two events can be together done in mn ways provided the events are independent.

npr = $$\frac{n !}{(n-r) !}$$ = n(n – 1)(n – 2) …………. (n – r + 1)

np1 = n, np2 = n(n-1), np3 = n(n-1)(n-2).

npr = n!

→ $$\frac{{ }^{\mathrm{n}} \mathrm{p}_{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{p}_{\mathrm{r}-1}}$$npn = n!

n+1pr = r. npr-1 + npr

npr = r. n-1pr-1 + npr

npr = r.n-1pr-1 + n-1pr

→ The number of permutations of n things taken r at a time containing a particular thing is r × n-1pr-1

→ The number of permutations of n things taken r at a time not containing a particular thing is n-1pr

• The number of permutations of n things taken r at a time allowing repetitions is nr.
• The number of permutations of n things taken not more than r at a time allowing repetitions is $$\frac{n\left(n^{\mathrm{r}}-1\right)}{(n-1)}$$

→ The number of permutations of n things of which p things are of one kind and q things are of another kind etc., is $$\frac{n !}{p ! q ! \ldots \ldots \cdots}$$

→ The sum of all possible numbers formed out of all the ‘n’ digits without zero is (n-1)! (sum of all the digits) (111 ……….. n times).

→ The sum of all possible numbers formed out of all the ‘n’ digits which includes zero is [(n -1)! (sum of all the digits) (111 …………… n times)] – [( n – 2)! (sum of all the digits)(111 ………….. (n -1) times]

→ The sum of all possible numbers formed by taking r digits from the given n digits which do not include zero is n-1pr-1 (sum of all the digits)(111 …………… r times).

→ The sum of all possible numbers formed by taking r digits from the given n digits which include zero is n-1pr-1(sum of all the digits)(111 ……………….. r times) – n-2pr-2(sum of all the digits)(111 ……………….. (r-1) times).

• The number of permutations of n things when arranged round a circle is (n -1)!
• In case of necklace or garland number of circular permutations is $$\frac{(n-1) !}{2}$$ → Number of permutations of n things taken r at a time in which there is at least one repetition is nrnpr.

→ The number of circular permutations of ‘n’ different things taken ‘r’ at a time is $$\frac{{ }^{n} \mathrm{P}_{\mathrm{r}}}{\mathrm{r}}$$