Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 11 Maths NCERT Solutions for the Chapter 6 Miscellaneous Exercise — all 11 questions solved, mixing permutations and combinations together the way CBSE tests them at exam level.
This exercise doesn't introduce new formulas — it tests whether the right one gets picked. Word-arrangement questions (1, 2, 4, 11) come back to permutations, often with a grouping condition layered on top. Selection questions (3, 6, 7, 8, 10) come back to combinations, frequently with an "at least" or "at most" condition that needs breaking into separate cases. Questions 5 and 9 mix a fixed constraint (a units digit, an even seat) with ordinary counting. Reading each question carefully for whether order matters is the real skill this exercise checks.
DAUGHTER has 8 letters — 3 vowels (A, U, E) and 5 consonants (D, G, H, T, R).
2 vowels can be chosen from the 3 vowels, and 3 consonants can be chosen from the 5 consonants:
{}^{3}C_{2}=3 \qquad {}^{5}C_{3}=10
Each such selection gives 5 letters in total, which can be arranged among themselves in 5!=120 ways.
By the multiplication principle, the required number of words is:
3\times10\times120=3600
EQUATION has 8 letters — 5 vowels (E, U, A, I, O) and 3 consonants (Q, T, N), all distinct.
Treating the vowels as one block and the consonants as another block gives 2 blocks to arrange, in 2!=2 ways.
Within the vowel block, the 5 distinct vowels can be arranged in 5!=120 ways. Within the consonant block, the 3 distinct consonants can be arranged in 3!=6 ways.
By the multiplication principle, the required number of words is:
2\times120\times6=1440
3 girls from 4, and the remaining 7-3=4 members from the 9 boys:
{}^{4}C_{3}\times{}^{9}C_{4}=4\times126=504
This means either exactly 3 girls, or exactly 4 girls (the maximum possible, since there are only 4 girls).
Exactly 3 girls: 504 ways (from part (i)).
Exactly 4 girls: all 4 girls, and the remaining 3 members from the 9 boys:
{}^{4}C_{4}\times{}^{9}C_{3}=1\times84=84
Adding the two cases:
504+84=588
This is every committee of 7 except the ones with all 4 girls. The total number of unrestricted committees of 7 from all 13 people is:
{}^{13}C_{7}=1716
Subtracting the "exactly 4 girls" case found in part (ii):
1716-84=1632
EXAMINATION has 11 letters: A (2), I (2), N (2), and E, X, M, T, O each occurring once.
In dictionary order, the letters of this word sorted alphabetically are A, A, E, I, I, M, N, N, O, T, X. Since A comes before E, every word that starts with A appears in the list before any word that starts with E.
So the required count is exactly the number of words starting with A. Fixing one A in the first position leaves 10 letters — E, X, A, M, I, N, T, I, O, N — to arrange in the remaining 10 places, where A now occurs once, I occurs twice, and N occurs twice.
\dfrac{10!}{2!\,2!}=\dfrac{3628800}{4}=907200
A number is divisible by 10 exactly when its units digit is 0, so the units place must be 0.
The remaining 5 places are filled by the remaining 5 digits (1, 3, 5, 7, 9), all non-zero, with no repetition — so there is no leading-digit restriction to worry about separately.
The number of ways to arrange these 5 digits in the remaining 5 places is:
5!=120
2 different vowels can be chosen from the 5 vowels, and 2 different consonants can be chosen from the 21 consonants:
{}^{5}C_{2}=10 \qquad {}^{21}C_{2}=210
Each such selection gives 4 distinct letters, which can be arranged among themselves in 4!=24 ways.
By the multiplication principle, the required number of words is:
10\times210\times24=50400
Let x questions be attempted from Part I (5 questions) and 8-x from Part II (7 questions), with x\ge3 and 8-x\ge3, i.e. 3\le x\le5 (since Part I has only 5 questions).
This gives three possible splits: 3 + 5, 4 + 4, and 5 + 3.
{}^{5}C_{3}\times{}^{7}C_{5}=10\times21=210
{}^{5}C_{4}\times{}^{7}C_{4}=5\times35=175
{}^{5}C_{5}\times{}^{7}C_{3}=1\times35=35
Adding all three cases:
210+175+35=420
A deck has 4 kings and 48 non-king cards. Exactly one king must be chosen from the 4 kings, and the remaining 4 cards of the hand must be chosen from the 48 non-king cards.
{}^{4}C_{1}=4 \qquad {}^{48}C_{4}=194580
By the multiplication principle, the required number of combinations is:
4\times194580=778320
A row of 9 seats has positions 1 to 9, of which the even positions are 2, 4, 6 and 8 — exactly 4 places, matching the 4 women.
The 4 women can be arranged in these 4 even places in 4!=24 ways.
The 5 men then fill the remaining 5 odd places (1, 3, 5, 7, 9) in 5!=120 ways.
By the multiplication principle, the required number of arrangements is:
24\times120=2880
The condition splits into two disjoint cases, based on whether the 3 particular students join together or not.
The remaining 10-3=7 members of the party must come from the other 25-3=22 students:
{}^{22}C_{7}=170544
All 10 members of the party must come from the other 22 students:
{}^{22}C_{10}=646646
Adding the two cases:
170544+646646=817190
ASSASSINATION has 13 letters: A (3), S (4), I (2), N (2), and T, O each occurring once.
Treating the block SSSS as a single unit, the objects to arrange become: [SSSS], A, A, A, I, I, N, N, T, O — a total of 10 objects, with A repeated 3 times, I repeated twice, and N repeated twice.
\dfrac{10!}{3!\,2!\,2!}=\dfrac{3628800}{6\times2\times2}=\dfrac{3628800}{24}=151200
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