NCERT Solutions · Class 11 Mathematics Permutations and Combinations

Fundamental principle of counting. The available digits are \(\displaystyle 1, 2, 3, \ldots, 9\) — nine digits, none of them \(\displaystyle 0\), so any arrangement of three of them is a valid $\displaystyle 3$-digit number.No digit may repeat, so fill the places in turn: \[9 \times 8 \times 7 = 504 \] This is the permutation \(\displaystyle {}^{9}P_{3} = \dfrac{9!}{6!} = 504\).$\displaystyle 504$ such $\displaystyle 3$-digit numbers can be formed.

Fill the leading place first. All ten digits \(\displaystyle 0, 1, 2, \ldots, 9\) are available, but a $\displaystyle 4$-digit number cannot begin with \(\displaystyle 0\) — otherwise it would be a $\displaystyle 3$-digit number. So the thousands place is the restricted one and must be settled first.Thousands place: any of \(\displaystyle 1, 2, \ldots, 9\), so \(\displaystyle 9\) ways.One digit is now used, leaving \(\displaystyle 9\) digits (including \(\displaystyle 0\)) for the other three places, with no repetition: \[9 \times 8 \times 7 = 504 \]By the fundamental principle of counting, \[9 \times 504 = 4536 \]There are $\displaystyle 4536$ four-digit numbers with no digit repeated.

Fill the restricted place first. The digits available are \(\displaystyle 1, 2, 3, 4, 6, 7\) — six digits, none of them \(\displaystyle 0\). A number is even precisely when its units digit is even, so the units place carries the condition.Units place: the even digits in the list are \(\displaystyle 2, 4, 6\), giving \(\displaystyle 3\) ways.No digit may repeat, so after the units digit is chosen, \(\displaystyle 5\) digits remain: \[3 \times 5 \times 4 = 60 \]$\displaystyle 60$ such $\displaystyle 3$-digit even numbers can be made.

Permutation, then the restricted place. Five digits \(\displaystyle 1, 2, 3, 4, 5\) are available and none may repeat, so a $\displaystyle 4$-digit number is an arrangement of \(\displaystyle 4\) of the \(\displaystyle 5\) digits. No digit is \(\displaystyle 0\), so there is no leading-digit restriction.Total count. \[{}^{5}P_{4} = \frac{5!}{(5-4)!} = \frac{120}{1} = 120 \] (directly: \(\displaystyle 5 \times 4 \times 3 \times 2 = 120\)).How many are even. Fill the units place first: it must be \(\displaystyle 2\) or \(\displaystyle 4\), so \(\displaystyle 2\) ways. Four digits then remain for the other three places: \[4 \times 3 \times 2 = 24 \] Hence the number of even ones is \[2 \times 24 = 48 \]$\displaystyle 120$ four-digit numbers in all, of which $\displaystyle 48$ are even.

Ordered selection — a permutation. The two posts are different: choosing person \(\displaystyle A\) as chairman and \(\displaystyle B\) as vice chairman is not the same appointment as \(\displaystyle B\) as chairman and \(\displaystyle A\) as vice chairman. So order matters, and one person cannot fill both posts.Chairman: any of the \(\displaystyle 8\) persons, so \(\displaystyle 8\) ways. Vice chairman: any of the remaining \(\displaystyle 7\), so \(\displaystyle 7\) ways. \[8 \times 7 = 56 \] That is \(\displaystyle {}^{8}P_{2} = \dfrac{8!}{6!} = 56\).The two posts can be filled in $\displaystyle 56$ ways.

Simplify the ratio before solving. Use \(\displaystyle {}^{n}P_{r} = \dfrac{n!}{(n-r)!}\): \[{}^{n-1}P_{3} = \frac{(n-1)!}{(n-1-3)!} = \frac{(n-1)!}{(n-4)!}, \qquad {}^{n}P_{4} = \frac{n!}{(n-4)!} \]Both have the same denominator \(\displaystyle (n-4)!\), so it cancels in the ratio: \[\frac{{}^{n-1}P_{3}}{{}^{n}P_{4}} = \frac{(n-1)!}{n!} = \frac{(n-1)!}{n \cdot (n-1)!} = \frac{1}{n} \]The given condition is therefore \[\frac{1}{n} = \frac{1}{9} \quad \Longrightarrow \quad n = 9 \]This is admissible, since \(\displaystyle {}^{n-1}P_{3}\) needs \(\displaystyle n - 1 \geq 3\) and \(\displaystyle {}^{n}P_{4}\) needs \(\displaystyle n \geq 4\).Check: \(\displaystyle {}^{8}P_{3} = 8 \times 7 \times 6 = 336\) and \(\displaystyle {}^{9}P_{4} = 9 \times 8 \times 7 \times 6 = 3024\), and \(\displaystyle 336 : 3024 = 1 : 9\).\(\displaystyle n = 9\).

Expand both permutations and cancel. Throughout, \(\displaystyle {}^{n}P_{r} = \dfrac{n!}{(n-r)!}\) with \(\displaystyle 0 \leq r \leq n\). Since \(\displaystyle {}^{5}P_{r}\) appears on the left, \(\displaystyle r\) must satisfy \(\displaystyle 0 \leq r \leq 5\) — this is what picks out the acceptable root at the end.(i) \(\displaystyle {}^{5}P_{r} = 2 \cdot {}^{6}P_{r-1}\). \[\frac{5!}{(5-r)!} = 2 \cdot \frac{6!}{\bigl(6-(r-1)\bigr)!} = 2 \cdot \frac{6!}{(7-r)!} \] Rearranging so the factorials sit together, \[\frac{(7-r)!}{(5-r)!} = 2 \cdot \frac{6!}{5!} = 2 \times 6 = 12 \] Now \(\displaystyle (7-r)! = (7-r)(6-r)(5-r)!\), so \(\displaystyle (5-r)!\) cancels: \[(7-r)(6-r) = 12 \] \[42 - 13r + r^{2} = 12 \quad \Longrightarrow \quad r^{2} - 13r + 30 = 0 \quad \Longrightarrow \quad (r-3)(r-10) = 0 \] So \(\displaystyle r = 3\) or \(\displaystyle r = 10\). But \(\displaystyle r \leq 5\), so \(\displaystyle r = 10\) is rejected.Check: \(\displaystyle {}^{5}P_{3} = 60\) and \(\displaystyle 2 \cdot {}^{6}P_{2} = 2 \times 30 = 60\).(ii) \(\displaystyle {}^{5}P_{r} = {}^{6}P_{r-1}\). The same working without the factor \(\displaystyle 2\): \[\frac{(7-r)!}{(5-r)!} = \frac{6!}{5!} = 6 \quad \Longrightarrow \quad (7-r)(6-r) = 6 \] \[r^{2} - 13r + 42 = 6 \quad \Longrightarrow \quad r^{2} - 13r + 36 = 0 \quad \Longrightarrow \quad (r-4)(r-9) = 0 \] So \(\displaystyle r = 4\) or \(\displaystyle r = 9\), and \(\displaystyle r = 9\) is rejected as \(\displaystyle r \leq 5\).Check: \(\displaystyle {}^{5}P_{4} = 120\) and \(\displaystyle {}^{6}P_{3} = 120\).(i) \(\displaystyle r = 3\) (ii) \(\displaystyle r = 4\).

Permutations of distinct letters. The word EQUATION has \(\displaystyle 8\) letters — E, Q, U, A, T, I, O, N — and no letter is repeated. The number of ways of arranging \(\displaystyle n\) distinct objects, all taken at a time, is \(\displaystyle n!\).\[{}^{8}\mathrm{P}_{8} = 8! = 8\times7\times6\times5\times4\times3\times2\times1 = 40320\]$\displaystyle 40320$ words can be formed.

Permutations of distinct letters. MONDAY has \(\displaystyle 6\) different letters — M, O, N, D, A, Y — and no letter may be repeated, so filling \(\displaystyle r\) places from them can be done in \(\displaystyle {}^{6}\mathrm{P}_{r}=\dfrac{6!}{(6-r)!}\) ways.(i) $\displaystyle 4$ letters at a time. \[{}^{6}\mathrm{P}_{4}=\frac{6!}{2!}=6\times5\times4\times3=360\](ii) All letters at a time. \[{}^{6}\mathrm{P}_{6}=6!=720\](iii) All letters, first letter a vowel. The vowels in MONDAY are O and A, so the first place can be filled in \(\displaystyle 2\) ways. The remaining \(\displaystyle 5\) letters then fill the other \(\displaystyle 5\) places in \(\displaystyle 5!\) ways. \[2\times5! = 2\times120 = 240\](i) $\displaystyle 360$ (ii) $\displaystyle 720$ (iii) $\displaystyle 240$

Complementary counting with repeated letters. MISSISSIPPI has \(\displaystyle 11\) letters: M once, I four times, S four times, P twice. When \(\displaystyle n\) letters contain repeats of \(\displaystyle p, q, \dots\) alike letters, the number of distinct arrangements is \(\displaystyle \dfrac{n!}{p!\,q!\cdots}\).Total distinct permutations: \[\frac{11!}{4!\,4!\,2!} = \frac{39916800}{24\times24\times2} = 34650\]Now count the ones in which the four I's do come together. Tie the four I's into a single block. The objects to arrange are that block together with M, S, S, S, S, P, P — that is \(\displaystyle 8\) objects, with S repeated \(\displaystyle 4\) times and P repeated \(\displaystyle 2\) times: \[\frac{8!}{4!\,2!} = \frac{40320}{48} = 840\] (The four I's are identical, so the block has only one internal ordering.)Subtract the "together" cases from the total: \[34650 - 840 = 33810\]In $\displaystyle 33810$ of the distinct permutations the four I's do not come together.

Arrangements with a repeated letter, under position constraints. PERMUTATIONS has \(\displaystyle 12\) letters: P, E, R, M, U, T, A, T, I, O, N, S. Only T is repeated (twice); every other letter occurs once. So a divisor of \(\displaystyle 2!\) appears in each count.(i) Words start with P and end with S. Those two places are fixed, leaving the \(\displaystyle 10\) letters E, R, M, U, T, A, T, I, O, N for the ten middle places, with T twice: \[\frac{10!}{2!} = \frac{3628800}{2} = 1814400\](ii) Vowels all together. The vowels are E, U, A, I, O — five distinct letters. Tie them into one block. That block together with the \(\displaystyle 7\) consonants P, R, M, T, T, N, S gives \(\displaystyle 8\) objects to arrange, T repeated twice: \[\frac{8!}{2!} = 20160\] Inside the block the \(\displaystyle 5\) vowels can be ordered in \(\displaystyle 5! = 120\) ways. \[20160\times120 = 2419200\](iii) Always $\displaystyle 4$ letters between P and S. P and S must occupy places whose numbers differ by \(\displaystyle 5\): \(\displaystyle (1,6), (2,7), (3,8), (4,9), (5,10), (6,11), (7,12)\) — that is \(\displaystyle 7\) pairs of places, and in each pair P and S can be interchanged, giving \(\displaystyle 7\times2 = 14\) choices. The other \(\displaystyle 10\) letters (T twice) fill the remaining ten places in \(\displaystyle \dfrac{10!}{2!} = 1814400\) ways. \[14\times1814400 = 25401600\](i) $\displaystyle 1814400$ (ii) $\displaystyle 2419200$ (iii) $\displaystyle 25401600$