Miscellaneous Exercise — Permutations and Combinations

1. 6.M.14 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

   How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?

   Hint. DAUGHTER has 3 vowels and 5 consonants. Select the required vowels and consonants first, then arrange the chosen 5 letters.

   DAUGHTER: vowels A,U,E (3), consonants D,G,H,T,R (5). Selecting 2 vowels from 3: 3C2=3. Selecting 3 consonants from 5: 5C3=10. Combinations: 3x10=30. Each group of 5 chosen letters can be arranged in 5!=120 ways. Total: 30x120=3600.

   ✦ Working through each part gives: 3600.

2. 6.M.24 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

   How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

   Hint. Treat the vowel group and the consonant group as two single blocks, arrange the blocks, then arrange each block's letters internally.

   EQUATION has 5 vowels (E,U,A,I,O) and 3 consonants (Q,T,N), all distinct. Treating each group as one block: 2 blocks arrange in 2! ways; the vowel block internally in 5! ways; the consonant block internally in 3! ways. Total: 2! x 5! x 3! = 2x120x6=1440.

   ✦ Working through each part gives: 1440.

3. 6.M.36 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

   A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (i) exactly 3 girls, (ii) at least 3 girls, (iii) at most 3 girls?

   Hint. For (ii) and (iii), list every valid split by number of girls (0 to 4) before computing.

   (i) Exactly 3 girls: 4C3 x 9C4 = 4x126=504. (ii) At least 3 girls: exactly 3 (504) plus exactly 4 (4C4x9C3=1x84=84): 504+84=588. (iii) At most 3 girls: sum of 0,1,2,3-girl cases: 4C0x9C7 + 4C1x9C6 + 4C2x9C5 + 4C3x9C4 = 36+336+756+504=1632.

   ✦ Working through each part gives: (i) 504. (ii) 588. (iii) 1632.

4. 6.M.45 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

   If the different permutations of all the letters of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?

   Hint. The only letter alphabetically before E among EXAMINATION's letters is A -- so count all words starting with A.

   EXAMINATION has 11 letters: A x2, I x2, N x2, and E,X,M,T,O once each. Words starting with A: fixing one A leaves 10 letters (A x1, I x2, N x2, plus E,X,M,T,O) to arrange: 10!/(2!2!) = 907200.

   ✦ Working through each part gives: 907200 words come before the first word starting with E.

5. 6.M.53 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

   How many 6-digit numbers can be formed from the digits 0,1,3,5,7 and 9 which are divisible by 10 and no digit is repeated?

   Hint. Divisibility by 10 fixes the last digit immediately, leaving only the remaining positions to arrange.

   Divisible by 10 means the units digit is 0. The remaining 5 digits (1,3,5,7,9) fill the other 5 positions: 5!=120.

   ✦ Working through each part gives: 120.

6. 6.M.64 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

   The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?

   Hint. Select the vowels and consonants independently, then arrange all 4 chosen letters.

   Selecting 2 vowels from 5: 5C2=10. Selecting 2 consonants from 21: 21C2=210. Combinations: 10x210=2100. Arranging the 4 chosen letters: 4!=24. Total: 2100x24=50400.

   ✦ Working through each part gives: 50400.

7. 6.M.75 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

   In an examination, a question paper consists of 12 questions divided into two parts, Part I and Part II, containing 5 and 7 questions respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

   Hint. List every valid split of 8 questions between the two parts that respects both the 'at least 3 from each' rule and each part's own size limit.

   Valid splits (Part I, Part II) summing to 8 with at least 3 from each and not exceeding each part's size (5 and 7): (3,5), (4,4), (5,3). Computing each: 5C3x7C5=10x21=210; 5C4x7C4=5x35=175; 5C5x7C3=1x35=35. Total: 210+175+35=420.

   ✦ Working through each part gives: 420.

8. 6.M.84 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

   Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

   Hint. Select the one king and the remaining four non-king cards independently, then multiply.

   1 king from 4: 4C1=4. Remaining 4 cards from the 48 non-king cards: 48C4=194580. Total: 4x194580=778320.

   ✦ Working through each part gives: 778320.

9. 6.M.94 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

   It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

   Hint. Count the even-numbered seats in a row of 9 first -- there are exactly as many as there are women.

   In a row of 9 seats, the even positions are 2,4,6,8 -- exactly 4 seats, matching the 4 women. The 4 women arrange in these 4 seats in 4!=24 ways. The remaining 5 odd seats (1,3,5,7,9) are filled by the 5 men in 5!=120 ways. Total: 24x120=2880.

   ✦ Working through each part gives: 2880.

10. 6.M.105 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

    From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?

    Hint. Split into two mutually exclusive cases based on whether the group of 3 joins or not, then add.

    Case 1: all 3 join -- choose the remaining 7 from the other 22 students: 22C7=170544. Case 2: none of the 3 join -- choose all 10 from the other 22 students: 22C10=646646. Total: 170544+646646=817190.

    ✦ Working through each part gives: 817190.

11. 6.M.115 marksNCERT Class 11 Mathematics, Permutations and Combinations, Reprint 2026-27

    In how many ways can the letters of the word ASSASSINATION be arranged so that all the S's are together?

    Hint. Treat the 4 S's as a single block, then arrange this block together with the remaining letters, accounting for any other repeated letters.

    ASSASSINATION has 13 letters: A x3, S x4, I x2, N x2, T x1, O x1. Treating the 4 S's as one block leaves 10 objects (block + A,A,A,I,I,N,N,T,O) to arrange, with A x3, I x2, N x2 among them: 10!/(3!2!2!) = 3628800/24=151200.

    ✦ Working through each part gives: 151200.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh106.pdf) — four numbered exercises (6.1-6.4, 31 questions) plus the chapter's Miscellaneous Exercise (11 questions); confirmed circular permutations do not appear anywhere in the chapter's theorems or its 42 questions, contrary to an earlier stub's claim about Exercise 6.3. Exercise 6.3 Q7's two-part nPr equation was solved with explicit domain checks (r must not exceed the smaller base number in each nPr term), rejecting an algebraically valid but combinatorially meaningless extraneous root in each part. Questions are referenced from the NCERT textbook for identification.