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 ⁿP_r and ⁿP_r = \(\frac{n !}{(n-r) !}\) for 0 ≤ r ≤ n.

→ If n, r positive integers and r ≤ n, then

• ⁿP_r = n. ^{n – 1}P_{r – 1} if r ≥ 1
• ⁿP_r = n.(n – 1) ^{n – 2}P_{r – 2} if r ≥ 2
• ⁿP_r = ^{n – 1}P_r + r. ^{(n – 1)}P_{(r – 1)}

→ The number of injections that can be defined from set A into set B is ^n(B)p_n(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 – 1}P_{r – 1}
• Not containing a particular thing is ^{n – 1}P_r
• Containing a particular thing in a particular place is ^{n – 1}P_{r – 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 n^r.

→ 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 ⁿC_r and ⁿC_r = \(\frac{n !}{(n-r) ! r !}\) for 0 ≤ r ≤ n and \(\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{n-r+1}{r}\) and ⁿP_r = r! ⁿC_r

→ If n, r are integers and 0 ≤ r ≤ n then ⁿc_r = ⁿC_{r – 1}

→ Let n, r, s are integers and 0 ≤ r ≤ n, 0 ≤ s ≤ n. If ⁿC_r = ⁿC_s 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)}C_m = ^{(m + n)}C_m = \(\frac{(m+n) !}{m ! n !}\)

→ 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 = p₁^α₁, p₁^α₂ …… p_k^α_k where p₁, p₂ …… p_k are distinct primes and α₁, α₂, …. α_k are non-negative integers, then the number of divisors of m is (α₁ + 1) (α₂ + 1) ……… (α_k + 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.

→ ⁿp_r 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.

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

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

→ ⁿp_r = n!

→ \(\frac{{ }^{\mathrm{n}} \mathrm{p}_{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{p}_{\mathrm{r}-1}}\)ⁿp_n = n!

→ ⁿ⁺¹p_r = r. ⁿp_r-1 + ⁿp_r

→ ⁿp_r = r. ^n-1p_r-1 + ⁿp_r

→ ⁿp_r = r.^n-1p_r-1 + ^n-1p_r

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

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

• The number of permutations of n things taken r at a time allowing repetitions is n^r.
• 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-1p_r-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-1p_r-1(sum of all the digits)(111 ……………….. r times) – ^n-2p_r-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 n^r – ⁿp_r.

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