Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 11 Maths NCERT Solutions for Chapter 6 Ex 6.3 — all 11 questions solved, covering permutations of digits, committee positions, equations in n and r, and word arrangements including EQUATION, MONDAY, MISSISSIPPI and PERMUTATIONS.
The formula {}^{n}P_{r}=\dfrac{n!}{(n-r)!} drives Questions 1–7 directly. From Question 8 onward, the exercise moves to arranging the letters of a whole word — straightforward when every letter is different (EQUATION, MONDAY), but requiring a division by p! for each repeated letter once a word like MISSISSIPPI or PERMUTATIONS is involved. The last two questions also add a grouping condition — letters kept together, or a fixed number of letters between two specific positions — which is best handled by treating the block as one unit first, then arranging inside it separately.
Since order matters and no digit repeats, the required number is the number of permutations of 9 different digits taken 3 at a time.
{}^{9}P_{3}=9\times8\times7=504
A 4-digit number cannot have 0 in the thousands (leftmost) place, so this place needs separate treatment from the rest.
Thousands place: can be filled by any of the 9 non-zero digits (1–9), so 9 ways.
The remaining 3 places are filled from the remaining 9 digits (0–9 minus the one already used), with no repetition — this is a permutation of 9 digits taken 3 at a time.
{}^{9}P_{3}=9\times8\times7=504
By the multiplication principle, the required number of 4-digit numbers is:
9\times504=4536
A number is even exactly when its units digit is even, so the units place must be filled first, from the restricted choices, before the unrestricted places.
Units place: only 2, 4 or 6 can go here — 3 ways.
Once the units digit is fixed, the tens and hundreds places are filled from the remaining 5 digits, with no repetition — a permutation of 5 digits taken 2 at a time.
{}^{5}P_{2}=5\times4=20
By the multiplication principle, the required number of 3-digit even numbers is:
20\times3=60
The required count is the number of permutations of 5 digits taken 4 at a time.
{}^{5}P_{4}=5\times4\times3\times2=120
The units digit must be even, so it can be filled only by 2 or 4 — 2 ways.
The remaining 3 places are filled from the remaining 4 digits, with no repetition:
{}^{4}P_{3}=4\times3\times2=24
By the multiplication principle, the number of even 4-digit numbers is:
24\times2=48
The chairman can be chosen from the 8 persons in 8 ways. Since one person cannot hold both positions, the vice chairman must then be chosen from the remaining 7 persons, in 7 ways.
By the multiplication principle, or equivalently, the number of permutations of 8 persons taken 2 at a time:
{}^{8}P_{2}=8\times7=56
Writing both permutations using the factorial formula:
{}^{n-1}P_{3}=\dfrac{(n-1)!}{(n-4)!} \qquad {}^{n}P_{4}=\dfrac{n!}{(n-4)!}
The given ratio becomes:
\dfrac{(n-1)!/(n-4)!}{n!/(n-4)!}=\dfrac{1}{9}
The (n-4)! cancels, and since n!=n\times(n-1)!:
\dfrac{(n-1)!}{n!}=\dfrac{1}{n}=\dfrac{1}{9}
So n=9.
Writing both sides using the factorial formula:
\dfrac{5!}{(5-r)!}=2\times\dfrac{6!}{(7-r)!}
Since (7-r)!=(7-r)(6-r)(5-r)!, cancelling (5-r)! from both sides:
5!=\dfrac{2\times6!}{(7-r)(6-r)}
(7-r)(6-r)=\dfrac{2\times6!}{5!}=2\times6=12
Expanding: 42-13r+r^2=12 \ \Rightarrow\ r^2-13r+30=0
(r-3)(r-10)=0 \ \Rightarrow\ r=3 \text{ or } r=10
Since r\le5 for {}^{5}P_{r} to be defined, r=10 is rejected.
Writing both sides using the factorial formula:
\dfrac{5!}{(5-r)!}=\dfrac{6!}{(7-r)!}
Cancelling (5-r)! as before:
(7-r)(6-r)=\dfrac{6!}{5!}=6
Expanding: 42-13r+r^2=6 \ \Rightarrow\ r^2-13r+36=0
(r-4)(r-9)=0 \ \Rightarrow\ r=4 \text{ or } r=9
Since r\le5, r=9 is rejected.
EQUATION has 8 letters — E, Q, U, A, T, I, O, N — all different, and every letter is used exactly once.
The required number of words is the number of permutations of 8 different letters taken all at a time:
8!=40320
MONDAY has 6 letters — M, O, N, D, A, Y — all different.
{}^{6}P_{4}=6\times5\times4\times3=360
{}^{6}P_{6}=6!=720
The vowels in MONDAY are O and A — 2 choices for the first letter.
The remaining 5 letters can be arranged in the remaining 5 places in 5! ways.
By the multiplication principle:
2\times5!=2\times120=240
MISSISSIPPI has 11 letters: M (1), I (4), S (4), P (2).
\dfrac{11!}{4!\,4!\,2!}=\dfrac{39916800}{24\times24\times2}=34650
Treat the block IIII as a single unit. This block, together with M, S, S, S, S, P, P, gives 8 objects in total, with S repeated 4 times and P repeated 2 times.
\dfrac{8!}{4!\,2!}=\dfrac{40320}{24\times2}=840
Subtract the "I's together" count from the total:
34650-840=33810
PERMUTATIONS has 12 letters, in which T appears twice and every other letter — P, E, R, M, U, A, I, O, N, S — appears once.
Fixing P at the first place and S at the last place leaves 10 letters (E, R, M, U, T, A, T, I, O, N) to arrange in the 10 middle places, with T repeated twice.
\dfrac{10!}{2!}=\dfrac{3628800}{2}=1814400
The vowels E, U, A, I, O (5 distinct vowels) are treated as a single block. This block, together with the 7 consonants P, R, M, T, T, N, S (T repeated twice), gives 8 objects to arrange.
\dfrac{8!}{2!}=\dfrac{40320}{2}=20160
Within the block, the 5 distinct vowels can themselves be arranged in 5!=120 ways.
By the multiplication principle, the required number of arrangements is:
20160\times120=2419200
Number the 12 positions 1 to 12. Exactly 4 letters lie between P and S when their positions differ by 5. The possible position pairs (smaller, larger) are (1,6), (2,7), (3,8), (4,9), (5,10), (6,11), (7,12) — 7 such pairs.
For each pair of positions, P and S can occupy them in 2 orders (P first, S second, or S first, P second), giving 7\times2=14 ways to place P and S.
The remaining 10 letters (E, R, M, U, T, A, T, I, O, N, with T repeated twice) fill the remaining 10 places in:
\dfrac{10!}{2!}=1814400 ways.
By the multiplication principle, the required number of arrangements is:
14\times1814400=25401600
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