NCERT Solutions · Class 11 Mathematics Permutations and Combinations

Fundamental principle of counting. A $\displaystyle 3$-digit number has three places — hundreds, tens and units — and each place must be filled by one of the digits \(\displaystyle 1, 2, 3, 4, 5\). None of these digits is \(\displaystyle 0\), so every filling gives a genuine $\displaystyle 3$-digit number.(i) Repetition allowed. The choice made at one place puts no restriction on the others, so each of the three places can be filled in \(\displaystyle 5\) ways. \[5 \times 5 \times 5 = 125 \](ii) Repetition not allowed. Fill the hundreds place in \(\displaystyle 5\) ways. One digit is now used up, so the tens place can be filled in \(\displaystyle 4\) ways, and the units place in \(\displaystyle 3\) ways. \[5 \times 4 \times 3 = 60 \] This is just \(\displaystyle {}^{5}P_{3}\).(i) $\displaystyle 125$ numbers, (ii) $\displaystyle 60$ numbers.

Fill the restricted place first. A number is even exactly when its units digit is even, so the units place is the one carrying a condition — settle it before the free places.Units place: the even digits available are \(\displaystyle 2, 4, 6\), so it can be filled in \(\displaystyle 3\) ways.Repetition is allowed, so the digit used at the units place is still available elsewhere. The tens place can be filled in \(\displaystyle 6\) ways and the hundreds place in \(\displaystyle 6\) ways (no digit is \(\displaystyle 0\), so the hundreds place needs no separate care).By the fundamental principle of counting, \[6 \times 6 \times 3 = 108 \]$\displaystyle 108$ such $\displaystyle 3$-digit even numbers can be formed.

Fundamental principle of counting. The first \(\displaystyle 10\) letters of the alphabet are \(\displaystyle a, b, c, \ldots, j\). A code is an ordered string of \(\displaystyle 4\) letters, and no letter may repeat.Fill the four positions one after another. The first can be any of the \(\displaystyle 10\) letters; the second any of the \(\displaystyle 9\) still unused; the third any of the remaining \(\displaystyle 8\); the fourth any of the remaining \(\displaystyle 7\). \[10 \times 9 \times 8 \times 7 = 5040 \] Equivalently \(\displaystyle {}^{10}P_{4} = \dfrac{10!}{6!} = 5040\).$\displaystyle 5040$ four-letter codes can be formed.

Fix the given digits, then count the rest. The number has \(\displaystyle 5\) places, and the first two are already decided: they must be \(\displaystyle 6\) and \(\displaystyle 7\). \[\underline{6}\ \ \underline{7}\ \ \underline{\phantom{x}}\ \ \underline{\phantom{x}}\ \ \underline{\phantom{x}} \] Since no digit may appear more than once, the digits \(\displaystyle 6\) and \(\displaystyle 7\) are now unavailable, leaving \(\displaystyle 10 - 2 = 8\) digits for the last three places.Third place: \(\displaystyle 8\) ways. Fourth place: \(\displaystyle 7\) ways. Fifth place: \(\displaystyle 6\) ways. \[8 \times 7 \times 6 = 336 \] That is \(\displaystyle {}^{8}P_{3} = 336\).$\displaystyle 336$ such telephone numbers can be constructed.

Multiplication principle. Each toss is recorded separately, and each toss has \(\displaystyle 2\) possible outcomes — head \(\displaystyle (\mathrm{H})\) or tail \(\displaystyle (\mathrm{T})\). The result of one toss does not restrict the next, so the three tosses can be combined freely. \[2 \times 2 \times 2 = 2^{3} = 8 \] The outcomes are \(\displaystyle \mathrm{HHH},\ \mathrm{HHT},\ \mathrm{HTH},\ \mathrm{HTT},\ \mathrm{THH},\ \mathrm{THT},\ \mathrm{TTH},\ \mathrm{TTT}\).There are $\displaystyle 8$ possible outcomes.

Ordered selection. A signal uses \(\displaystyle 2\) of the \(\displaystyle 5\) flags, one placed below the other. The two positions — upper and lower — are distinguishable, so a red flag above a blue flag is a different signal from a blue flag above a red one. Order matters, so this is a permutation.Upper position: any of the \(\displaystyle 5\) flags, so \(\displaystyle 5\) ways. Lower position: that flag is used, so \(\displaystyle 4\) ways. \[5 \times 4 = 20 \] That is \(\displaystyle {}^{5}P_{2} = \dfrac{5!}{3!} = 20\).$\displaystyle 20$ different signals can be generated.