JEE Mathematics Permutation and Combination MCQs Set 19

Answer: (b) \( {}^{11}P_{11}-1 \)

Explanation:
1. \( r\cdot r!=(r+1)!-r! \).
2. The sum telescopes: \( \sum_{r=1}^{10}\left[(r+1)!-r!\right]=11!-1! \).
3. So the value is \( 11!-1={}^{11}P_{11}-1 \).

Teacher's Note:
1. Write r times r factorial as a difference of consecutive factorials.
2. Then the series telescopes.

Answer: (a) 13

Explanation:
1. \( {}^{n}C_5={}^{n}C_8 \) with \( 5\ne8 \) means \( 5+8=n \).
2. So \( n=13 \).

Teacher's Note:
1. If nCa = nCb and a is not b, then a + b = n.

Answer: (c) \( 9\,(9!) \)

Explanation:
1. The first digit can be any of 1 to 9: 9 ways.
2. The remaining 8 places are filled from the 9 digits left: \( {}^{9}P_8=9! \).
3. Total \( 9\times9! \).

Teacher's Note:
1. The first digit cannot be zero.
2. Fill the first place before the others.

Answer: (b) \( {}^{4}P_4\times{}^{3}P_3 \)

Explanation:
1. Treat the three particular men as one block, giving 4 objects to arrange: \( 4!={}^{4}P_4 \).
2. The three men can be arranged within the block in \( 3!={}^{3}P_3 \) ways.
3. Total \( {}^{4}P_4\times{}^{3}P_3=144 \).

Teacher's Note:
1. Glue the men together as one unit, then multiply by the internal arrangements.

Answer: (d) none of these

Explanation:
1. All arrangements of 7 lectures: \( 7!=5040 \).
2. The three lectures A, B, C can be in any of \( 3!=6 \) orders and only one order is allowed.
3. Required number \( =\dfrac{7!}{3!}=840 \), which is not among (a), (b) and (c).

Teacher's Note:
1. For a fixed order among k objects, divide the total arrangements by k factorial.

Answer: (a) 216

Explanation:
1. Five of the six digits are used; the digit left out must make the digit sum a multiple of 3.
2. Total of all six digits is 15, so the left-out digit must be 0 or 3.
3. Leave out 0: digits 1, 2, 3, 4, 5 give \( 5!=120 \). Leave out 3: digits 0, 1, 2, 4, 5 give \( 5!-4!=96 \).
4. Total \( 120+96=216 \).

Teacher's Note:
1. A number is divisible by 3 when its digit sum is.
2. Remove arrangements starting with zero when 0 is used.

Answer: (c) 35

Explanation:
1. Place the six '+' signs first; they create 7 gaps (including both ends).
2. Choose 4 of these gaps for the '-' signs: \( {}^{7}C_4=35 \).

Teacher's Note:
1. Arrange the unrestricted objects first, then insert the restricted ones in the gaps.
2. Identical signs are not arranged among themselves.

Answer: (a) 384

Explanation:
1. First place: 1, 2 or 3, so 3 ways.
2. The three middle places: 4 ways each, \( 4^{3}=64 \).
3. Last place must be even: 0 or 2, so 2 ways.
4. Total \( 3\times64\times2=384 \).

Teacher's Note:
1. Handle the first and last places first because they are restricted.
2. Repetition allowed means each middle place has all four choices.

Answer: (b) 505

Explanation:
1. Total 4-digit numbers with repetition allowed: \( 5^{4}=625 \).
2. Numbers with all digits different: \( {}^{5}P_4=120 \).
3. At least two identical: \( 625-120=505 \).

Teacher's Note:
1. Use the complement: total minus numbers with all digits different.

Answer: (a) \( \dfrac{5(7!)}{2} \)

Explanation:
1. The nine letters are C, C, A, A, L, L, U, T, E, so \( \dfrac{9!}{2!\,2!\,2!}=45360 \) arrangements in all.
2. By symmetry, the first letter is a consonant with probability \( \dfrac{5}{9} \) (five consonants among nine letters), and then the last letter is a consonant with probability \( \dfrac{4}{8} \).
3. Required number \( =45360\times\dfrac{5}{9}\times\dfrac{4}{8}=12600=\dfrac{5(7!)}{2} \).

Teacher's Note:
1. Fix the end letters first, then arrange the rest allowing for repeats.
2. Treat each combination of end letters as a separate case.

Answer: (b) 84

Explanation:
1. The digits strictly increase from left to right, and the first digit is at least 1, so zero cannot be used.
2. Choose any 6 different digits from 1 to 9; each choice gives exactly one number.
3. \( {}^{9}C_6=84 \).

Teacher's Note:
1. Strictly increasing digits means the order is forced by the choice.
2. Zero would have to be first, which is not allowed.