Class 11 Maths Permutation & Combination and Binomial Theorem Test 1 and Step by Step Solutions – 2027

Below are the detailed, step-by-step solutions to the Practice Test on Permutation & Combination and Binomial Theorem. The test carries a maximum of 30 marks and is designed to be completed in 60 minutes. This test should be attempted after completing the chapter as a summative assessment exercise.
Designed for both students and teachers, these answers make exam revision easier with clear explanations, formula highlights, and downloadable PDF options.

Detailed Step-by-Step Solutions

Section A: 1 Mark Questions

Q1. Convert the following product into factorials: 5.6.7.8.9.10

Solution:
The given product is 5 \times 6 \times 7 \times 8 \times 9 \times 10.
To express this as a factorial, we can multiply and divide the expression by 4! (which expands to 1 \times 2 \times 3 \times 4).

\frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10}{1 \times 2 \times 3 \times 4}

This simplifies perfectly to: \frac{10!}{4!}

Final Answer: \boxed {\frac{10!}{4!}}

Q2. In how many ways can 7 letters be posted in 4 letter boxes?

Solution:
We have 7 letters to be posted into 4 letter boxes.
Each of the 7 letters has 4 independent options (any of the 4 available letter boxes).

Therefore, the total number of ways is: 4 \times 4 \times 4 \times 4 \times 4 \times 4 \times 4 = 4^7

Final Answer: \boxed {4^7}

Q3. How many terms are there in the expansion of 25x^2 + 30xy + 9y^2 raised to the power of 11?

Solution:
First, observe the inner expression 25x^2 + 30xy + 9y^2.
By factorization, this is a perfect square: (5x + 3y)^2.

Substituting this back into the original expression, we get:
((5x + 3y)^2)^{11} = (5x + 3y)^{22}

For any standard binomial expansion (a + b)^n, the total number of terms is n + 1.
Thus, the number of terms is 22 + 1 = 23.

Final Answer: \boxed{23}

Section B: 2 Marks Questions

Q4. How many words with or without meaning, can be formed by using all the letters of the word ‘DELHI’, using each letter exactly once?

Solution:
The word ‘DELHI’ consists of 5 distinct letters.
Since we must use each letter exactly once, we are simply arranging 5 items in 5 places.

Total number of words = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120.

Final Answer: \boxed{120}

Q5. How many words can be formed out of the letters of the word ‘ARTICLE’, so that vowels occupy even places?

Solution:
The word ‘ARTICLE’ has 7 letters in total.
Vowels: A, I, E (3 vowels)
Consonants: R, T, C, L (4 consonants)

In a 7-letter word, the even places are the 2nd, 4th, and 6th positions, giving us exactly 3 even places.

Ways to arrange the 3 vowels in the 3 even places = 3! = 6.
Ways to arrange the remaining 4 consonants in the remaining 4 places = 4! = 24.

Total words formed = 3! \times 4! = 6 \times 24 = 144.

Final Answer: \boxed{144 }

Q6. Using Binomial theorem evaluate (9999)^5

Solution:
To make the calculation manageable, we can rewrite 9999 as 10000 - 1,
which can be expressed in exponential form as (10^4 - 1).

This allows us to apply the Binomial Theorem.
According to the Binomial Theorem for (a - b)^n: (a - b)^n = \binom{n}{0}a^n - \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 - \dots + (-1)^n \binom{n}{n}b^n

Applying this formula to our expression (10^4 - 1)^4, where a = 10^4, b = 1, and n = 4:

(10^4 - 1)^4 = \binom{4}{0}(10^4)^4 - \binom{4}{1}(10^4)^3(1) + \binom{4}{2}(10^4)^2(1)^2 - \binom{4}{3}(10^4)^1(1)^3 + \binom{4}{4}(10^4)^0(1)^4

Now, let’s expand the binomial coefficients (\binom{4}{0}=1, \binom{4}{1}=4, \binom{4}{2}=6, \binom{4}{3}=4, \binom{4}{4}=1) and multiply the powers of 10:

= 1(10^{16}) - 4(10^{12}) + 6(10^8) - 4(10^4) + 1(1)
= 10,000,000,000,000,000 - 4,000,000,000,000 + 600,000,000 - 40,000 + 1

To avoid calculation errors (which we know students often make in the exam hall!), it is best to group the positive and negative terms together before finalizing:

Sum of Positive terms:
10,000,000,000,000,000 + 600,000,000 + 1 = 10,000,000,600,000,001

Sum of Negative terms:
4,000,000,000,000 + 40,000 = 4,000,000,040,000

Finally, subtract the total negative sum from the total positive sum:
10,000,000,600,000,001 - 4,000,000,040,000 = 9,996,000,599,960,001

Final Answer: \boxed{ (9999)^4 = 9,996,000,599,960,001}

Q7. How many permutations of the letters of the word ‘MADHUBANI’ do not begin with M but end with I?

Solution:
The word ‘MADHUBANI’ has 9 letters in total, with the letter ‘A’ repeating twice and all other letters occurring exactly once.
Let’s break this down into conditions:

Condition 1: The word must end with I. Let’s fix ‘I’ at the last position.
The remaining 8 letters (M, A, A, D, H, U, B, N) can be arranged in \frac{8!}{2!} = 20,160 ways.

Condition 2: The word must NOT begin with M.
From our total permutations ending in ‘I’, we must subtract the specific permutations that begin with ‘M’ and end with ‘I’.
By fixing ‘M’ at the first position and ‘I’ at the last, we are left with 7 letters (A, A, D, H, U, B, N). These can be arranged in \frac{7!}{2!} = 2,520 ways.

Required permutations = (Total words ending in I) – (Words beginning with M and ending in I)
20,160 - 2,520 = 17,640

Final Answer: \boxed {17,640}

Q8. If C^{2n}_3 : C^n_2 = 44: 3, find n.

Solution:
We are given the combination ratio \frac{^{2n}C_3}{^nC_2} = \frac{44}{3}.

Applying the combination formula:
\frac{\frac{2n(2n-1)(2n-2)}{3 \times 2 \times 1}}{\frac{n(n-1)}{2 \times 1}} = \frac{44}{3}

\frac{2n(2n-1)2(n-1)}{6} \times \frac{2}{n(n-1)} = \frac{44}{3}

Canceling out the common n(n-1) terms from the numerator and denominator:
\frac{8(2n-1)}{6} = \frac{44}{3}

\frac{4(2n-1)}{3} = \frac{44}{3}

Cross-multiplying or simply canceling the denominator 3:
4(2n-1) = 44 \implies 2n-1 = 11 \implies 2n = 12 \implies n = 6

Final Answer: \boxed{ n=6 }

Q9. In how many ways can a committee of 5 persons be formed out of 6 men and 4 women when at least one woman has to be necessarily selected?

Solution:
We have a total of 10 people (6 men + 4 women), and we need to form a committee of 5 persons.

Total ways to select 5 people strictly without any restrictions = ^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252.

Ways to form a committee with NO women (meaning all 5 selected are men) = ^6C_5 = 6.

Since at least one woman must necessarily be selected, we can subtract the “no women” case from the overall unrestricted cases:
252 - 6 = 246 \text{ ways}

Final Answer: \boxed{ 246 }

Q10. Find the term independent of x in the expansion of (\frac{3}{2}x^2 - \frac{1}{3x})^9

Solution:
For the binomial expansion of (\frac{3}{2}x^2 - \frac{1}{3x})^9,

the general term T_{r+1} is represented as:
T_{r+1} = \binom{9}{r} \left(\frac{3}{2}x^2\right)^{9-r} \left(-\frac{1}{3x}\right)^r

= \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r x^{2(9-r) - r}

= \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r x^{18 - 3r}

For the term to be completely independent of x, the exponent of x must be 0:
18 - 3r = 0 \implies 3r = 18 \implies r = 6

Now, substitute r = 6 back into the coefficient part of our general term:
T_7 = \binom{9}{6} \left(\frac{3}{2}\right)^3 \left(-\frac{1}{3}\right)^6 = 84 \times \frac{27}{8} \times \frac{1}{729}

T_7 = 84 \times \frac{1}{8 \times 27} = \frac{84}{216} = \frac{7}{18}

Final Answer: \boxed { T_7 = \frac{7}{18}}

Q11. Using binomial theorem write the expansion of (\frac{x}{a} - \frac{a}{x})^6

Solution: Applying the standard Binomial Theorem:
\left(\frac{x}{a} - \frac{a}{x}\right)^6 = \binom{6}{0}\left(\frac{x}{a}\right)^6 - \binom{6}{1}\left(\frac{x}{a}\right)^5\left(\frac{a}{x}\right) + \binom{6}{2}\left(\frac{x}{a}\right)^4\left(\frac{a}{x}\right)^2 - \binom{6}{3}\left(\frac{x}{a}\right)^3\left(\frac{a}{x}\right)^3 + \binom{6}{4}\left(\frac{x}{a}\right)^2\left(\frac{a}{x}\right)^4 - \binom{6}{5}\left(\frac{x}{a}\right)\left(\frac{a}{x}\right)^5 + \binom{6}{6}\left(\frac{a}{x}\right)^6

Simplifying the powers cancels variables nicely:
\left(\frac{x}{a} - \frac{a}{x}\right)^6 = \frac{x^6}{a^6} - 6\left(\frac{x^4}{a^4}\right) + 15\left(\frac{x^2}{a^2}\right) - 20 + 15\left(\frac{a^2}{x^2}\right) - 6\left(\frac{a^4}{x^4}\right) + \frac{a^6}{x^6}

Q12. There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated

Solution:
The hall is equipped with 10 independently operated lamps.
Each lamp has exactly 2 possible states: ON or OFF.
Total possible switch combinations for all 10 lamps = 2 \times 2 \times \dots \text{ (10 times)} = 2^{10} = 1024.

The hall will be illuminated as long as at least one lamp is switched ON.
The only specific scenario where the hall remains entirely dark is when ALL lamps are turned OFF (1 way).
Therefore, the number of ways the hall can be illuminated = Total combinations – All OFF = 1024 - 1 = 1023.

Final Answer: \boxed{ 1023 }

Section C: 3 Marks Questions

Q13. If the letters of the word ‘MOTHER’ are written in all possible orders and these words are written out as in a dictionary. Find the rank of the word ‘MOTHER’.

Solution:
To find dictionary rank, we first arrange the letters of ‘MOTHER’ in alphabetical order: E, H, M, O, R, T.
There are 6 letters in total.

• Words naturally starting with E = 5! = 120
• Words naturally starting with H = 5! = 120
• Words starting with M… Since our target word ‘MOTHER’ starts with M, we freeze ‘M’ and evaluate the next available letter. Remaining letters are E, H, O, R, T.
  • Words starting with ME = 4! = 24
  • Words starting with MH = 4! = 24
  • Words starting with MO… Correct for ‘MOTHER’, so we freeze ‘O’. Remaining letters are E, H, R, T.
    • Words starting with MOE = 3! = 6
    • Words starting with MOH = 3! = 6
    • Words starting with MOR = 3! = 6
    • Words starting with MOT… Correct, so we freeze ‘T’. Remaining are E, H, R.
      • Words starting with MOTE = 2! = 2
      • Words starting with MOTH… Correct, so freeze ‘H’. Remaining are E, R.
        • Words starting with MOTHE… Correct, so freeze ‘E’. The only remaining letter is R.
          • The very next word formed is exactly ‘MOTHER’. This counts as 1 word.

Summing up the arrangements: 120 + 120 + 24 + 24 + 6 + 6 + 6 + 2 + 1 = 309.
The rank of the word ‘MOTHER’ is 309.

Final Answer: \boxed { 309 }

Q14. How many words can be formed by taking 4 letters at a time from the letters of the word ‘MORADABAD’

Solution:
The given word ‘MORADABAD’ contains 9 letters total. Let’s group them by frequency:
A appears 3 times.
D appears 2 times.
M, O, R, B appear 1 time each.

This gives us 6 distinct letter options: {A, D, M, O, R, B}. Since we need to form 4-letter words, we must approach this by breaking it into 4 mutually exclusive cases:

Case 1: All 4 letters selected are distinct.
Choose 4 letters out of the 6 distinct types: ^6C_4 = 15.
Ways to arrange them = 15 \times 4! = 15 \times 24 = 360.

Case 2: 2 alike of one kind, and 2 distinct letters.
Choose 1 pair from the 2 available letter pairs (A or D): ^2C_1 = 2.
Choose 2 distinct letters from the remaining 5 available types: ^5C_2 = 10.
Ways to arrange them = (2 \times 10) \times \frac{4!}{2!} = 20 \times 12 = 240.

Case 3: 2 alike of one kind, and 2 alike of another kind.
Choose 2 pairs from the 2 available pairs (A and D): ^2C_2 = 1.
Ways to arrange them = 1 \times \frac{4!}{2!2!} = 1 \times 6 = 6.

Case 4: 3 alike of one kind, and 1 distinct letter.
Choose 1 triplet from the 1 available triplet (A): ^1C_1 = 1.
Choose 1 distinct letter from the remaining 5 available types: ^5C_1 = 5.
Ways to arrange them = (1 \times 5) \times \frac{4!}{3!} = 5 \times 4 = 20.

Total 4-letter words formed = 360 + 240 + 6 + 20 = 626.

Final Answer: Total number of 4 letter words: \boxed { 626 }

Q15. Using Binomial Theorem, prove that 2^{3n} - 7n - 1 is divisible by 49, where n \in N.

Solution:
First, we apply exponent rules to rewrite 2^{3n} as (2^3)^n = 8^n.
We can further express 8 mathematically as 1 + 7.

So, 8^n = (1 + 7)^n.

Now, let’s expand (1 + 7)^n using the Binomial Theorem:
(1 + 7)^n = \binom{n}{0}1^n + \binom{n}{1}7 + \binom{n}{2}7^2 + \binom{n}{3}7^3 + \dots + \binom{n}{n}7^n

Substituting the basic binomial coefficients:
8^n = 1 + 7n + \binom{n}{2}49 + \binom{n}{3}343 + \dots + 7^n

Subtract 1[latex] and [latex]7n from both sides to mirror our required expression:
8^n - 1 - 7n = 49\left[ \binom{n}{2} + \binom{n}{3}7 + \dots + \binom{n}{n}7^{n-2} \right]

Because n is a natural number (n \in N), the entire expression grouped inside the square brackets will result in an integer.

Let's represent this integer as k.
2^{3n} - 7n - 1 = 49k

Because 2^{3n} - 7n - 1 perfectly equals a multiple of 49, it is exactly divisible by 49. Hence proved.