NCERT Solutions · Class 11 Mathematics Permutations and Combinations

Select, then arrange. DAUGHTER has \(\displaystyle 8\) distinct letters: the vowels A, U, E (three of them) and the consonants D, G, H, T, R (five of them).Selecting the letters — $\displaystyle 2$ vowels from $\displaystyle 3$ and $\displaystyle 3$ consonants from $\displaystyle 5$: \[{}^{3}\mathrm{C}_{2}\times{}^{5}\mathrm{C}_{3} = 3\times10 = 30\]Arranging each such choice — every selection has $\displaystyle 5$ different letters, so it gives \[5! = 120\] words.\[30\times120 = 3600\]$\displaystyle 3600$ words.

Group-and-arrange method. EQUATION has \(\displaystyle 8\) distinct letters: the vowels A, E, I, O, U (five) and the consonants Q, T, N (three). Requiring the vowels to occur together and the consonants to occur together means all five vowels form one unbroken block and all three consonants form another.The two blocks can be placed in \(\displaystyle 2!\) orders (vowel block first, or consonant block first). Inside the blocks, the vowels can be ordered in \(\displaystyle 5!\) ways and the consonants in \(\displaystyle 3!\) ways.\[2!\times5!\times3! = 2\times120\times6 = 1440\]$\displaystyle 1440$ words.

Case-split on the number of girls. A committee of \(\displaystyle 7\) is chosen from \(\displaystyle 9\) boys and \(\displaystyle 4\) girls. Note that only \(\displaystyle 4\) girls exist, so a committee can hold at most $\displaystyle 4$ of them.(i) Exactly $\displaystyle 3$ girls — $\displaystyle 3$ girls and \(\displaystyle 7-3=4\) boys: \[{}^{4}\mathrm{C}_{3}\times{}^{9}\mathrm{C}_{4} = 4\times126 = 504\](ii) At least $\displaystyle 3$ girls — $\displaystyle 3$ girls or $\displaystyle 4$ girls: \[\underbrace{{}^{4}\mathrm{C}_{3}\times{}^{9}\mathrm{C}_{4}}_{504} + \underbrace{{}^{4}\mathrm{C}_{4}\times{}^{9}\mathrm{C}_{3}}_{1\times84\,=\,84} = 588\](iii) At most $\displaystyle 3$ girls — $\displaystyle 0$, $\displaystyle 1$, $\displaystyle 2$ or $\displaystyle 3$ girls: \[{}^{4}\mathrm{C}_{0}\times{}^{9}\mathrm{C}_{7} = 1\times36 = 36\] \[{}^{4}\mathrm{C}_{1}\times{}^{9}\mathrm{C}_{6} = 4\times84 = 336\] \[{}^{4}\mathrm{C}_{2}\times{}^{9}\mathrm{C}_{5} = 6\times126 = 756\] \[{}^{4}\mathrm{C}_{3}\times{}^{9}\mathrm{C}_{4} = 4\times126 = 504\] \[36+336+756+504 = 1632\] Check: all committees number \(\displaystyle {}^{13}\mathrm{C}_{7}=1716\), and removing the only excluded case ($\displaystyle 4$ girls, $\displaystyle 84$ committees) leaves \(\displaystyle 1716-84=1632\).(i) $\displaystyle 504$ (ii) $\displaystyle 588$ (iii) $\displaystyle 1632$

Count the whole alphabetical block that precedes the E-words. EXAMINATION has \(\displaystyle 11\) letters: A twice, I twice, N twice, and E, X, M, T, O once each.In dictionary order the distinct first letters, listed increasingly, are A, E, I, M, N, O, T, X. So every word listed before the first word beginning with E is exactly a word beginning with A, and all of them come before it.Fix one A in the first place. The remaining \(\displaystyle 10\) letters are A, E, I, I, M, N, N, O, T, X, in which I occurs twice and N occurs twice: \[\frac{10!}{2!\,2!} = \frac{3628800}{4} = 907200\]$\displaystyle 907200$ words appear in the list before the first word starting with E.

Fix the forced digit first. A number is divisible by \(\displaystyle 10\) exactly when its units digit is \(\displaystyle 0\). A $\displaystyle 6$-digit number made from the six digits \(\displaystyle 0, 1, 3, 5, 7, 9\) with no repetition must use each digit once, so the \(\displaystyle 0\) is forced into the units place and can go nowhere else.The remaining five digits \(\displaystyle 1, 3, 5, 7, 9\) then fill the other five places in \[5! = 5\times4\times3\times2\times1 = 120\] ways. Since \(\displaystyle 0\) is already used up at the end, the leading digit is automatically non-zero, so every one of these arrangements really is a $\displaystyle 6$-digit number.$\displaystyle 120$ such numbers.

Choose, then arrange.A "word" here is any arrangement of the four chosen letters, meaningful or not. Building one takes three independent decisions, so the answer is the product of the three counts.Step $\displaystyle 1$ — choose the vowels. Order does not matter while selecting, so this is a combination: \[{}^{5}C_{2}=\frac{5!}{2!\,3!}=\frac{5\times 4}{2}=10 \]Step $\displaystyle 2$ — choose the consonants. \[{}^{21}C_{2}=\frac{21!}{2!\,19!}=\frac{21\times 20}{2}=210 \]Step $\displaystyle 3$ — arrange the $\displaystyle 4$ chosen letters. They are all different, so they can be placed in the four positions in \(\displaystyle 4!=24\) ways.By the fundamental principle of counting, \[10\times 210\times 24=50400 \]$\displaystyle 50400$ words can be formed.

Case analysis on the split between the parts.Let the student take \(\displaystyle a\) questions from Part I (which has $\displaystyle 5$) and \(\displaystyle b\) from Part II (which has $\displaystyle 7$), with \(\displaystyle a+b=8\), \(\displaystyle a\ge 3\), \(\displaystyle b\ge 3\).From \(\displaystyle b=8-a\ge 3\) we get \(\displaystyle a\le 5\), and \(\displaystyle a\le 5\) anyway since Part I has only $\displaystyle 5$ questions. So \(\displaystyle a\in\{3,4,5\}\) — exactly three cases. Selecting questions is a combination, since order of attempt is irrelevant.

Part I \(\displaystyle (a)\)	Part II \(\displaystyle (b)\)	Number of ways
$\displaystyle 3$	$\displaystyle 5$	\(\displaystyle {}^{5}C_{3}\times{}^{7}C_{5}=10\times 21=210\)
$\displaystyle 4$	$\displaystyle 4$	\(\displaystyle {}^{5}C_{4}\times{}^{7}C_{4}=5\times 35=175\)
$\displaystyle 5$	$\displaystyle 3$	\(\displaystyle {}^{5}C_{5}\times{}^{7}C_{3}=1\times 35=35\)

The three cases are mutually exclusive, so add them: \[210+175+35=420 \]The student can select the questions in $\displaystyle 420$ ways.

Split the deck into kings and non-kings.A pack of $\displaystyle 52$ cards contains $\displaystyle 4$ kings and \(\displaystyle 52-4=48\) other cards. "Exactly one king" means the hand takes $\displaystyle 1$ card from the $\displaystyle 4$ kings and the remaining $\displaystyle 4$ cards from the $\displaystyle 48$ non-kings.The king: \[{}^{4}C_{1}=4 \]The other four cards: \[{}^{48}C_{4}=\frac{48\times 47\times 46\times 45}{4\times 3\times 2\times 1}=194580 \]The two choices are independent, so multiply: \[4\times 194580=778320 \]There are $\displaystyle 778320$ such $\displaystyle 5$-card combinations.

Fix the positions first, then permute within them.There are \(\displaystyle 5+4=9\) people, so the row has $\displaystyle 9$ seats numbered $\displaystyle 1$ to 9.The even-numbered seats are $\displaystyle 2$, $\displaystyle 4$, $\displaystyle 6$, $\displaystyle 8$ — exactly $\displaystyle 4$ of them, which is precisely the number of women. So the women must fill seats $\displaystyle 2$, $\displaystyle 4$, $\displaystyle 6$, $\displaystyle 8$ and the men are forced into the five odd seats $\displaystyle 1$, $\displaystyle 3$, $\displaystyle 5$, $\displaystyle 7$, 9. There is no choice about which seats each group gets; only the order within a group varies.Women among the $\displaystyle 4$ even seats: \(\displaystyle 4!=24\) ways.Men among the $\displaystyle 5$ odd seats: \(\displaystyle 5!=120\) ways.These are independent, so \[4!\times 5!=24\times 120=2880 \]$\displaystyle 2880$ arrangements are possible.

Two exclusive cases: the trio is in, or the trio is out.Apart from the $\displaystyle 3$ students who move as a block, there are \(\displaystyle 25-3=22\) other students.Case $\displaystyle 1$ — all $\displaystyle 3$ join. They occupy $\displaystyle 3$ of the $\displaystyle 10$ places, so the remaining \(\displaystyle 10-3=7\) members come from the other $\displaystyle 22$: \[{}^{22}C_{7}=170544 \]Case $\displaystyle 2$ — none of the $\displaystyle 3$ joins. All $\displaystyle 10$ members come from the other $\displaystyle 22$: \[{}^{22}C_{10}=646646 \]No party can satisfy both conditions at once, so add the cases: \[170544+646646=817190 \]The party can be chosen in \(\displaystyle {}^{22}C_{7}+{}^{22}C_{10}=817190\) ways.

Tie the block together, then use the repeated-letters formula.ASSASSINATION has $\displaystyle 13$ letters: A appears $\displaystyle 3$ times, S appears $\displaystyle 4$ times, I appears $\displaystyle 2$ times, N appears $\displaystyle 2$ times, and T, O appear once each \(\displaystyle (3+4+2+2+1+1=13)\).Since all four S's must stay together, glue them into a single block \(\displaystyle [\text{SSSS}]\). We are now arranging \[\text{A, A, A, I, I, N, N, T, O and the block} \] which is \(\displaystyle 9+1=10\) objects in a row.Among these $\displaystyle 10$ objects, A repeats $\displaystyle 3$ times, I repeats $\displaystyle 2$ times and N repeats $\displaystyle 2$ times. The number of arrangements of \(\displaystyle n\) objects of which \(\displaystyle p_1,p_2,\dots\) are alike is \(\displaystyle \dfrac{n!}{p_1!\,p_2!\cdots}\), so \[\frac{10!}{3!\,2!\,2!}=\frac{3628800}{6\times 2\times 2}=\frac{3628800}{24}=151200 \]The four S's are identical, so permuting them inside the block gives nothing new — no extra factor is needed.$\displaystyle 151200$ arrangements have all the S's together.