34. If the different permutations of all the letter of the word EXAMINATION arelisted as in a dictionary, how many words are there in this list before the firstword starting with E ?

34. In the eleven-letter word EXAMINATION there are 2A's, 2I's and 2N's and the rest are all different.

Since in a dictionary the words are listed according to the English alphabet we can only find words starting with A (as B, C, D are not a part of the letters forming the word EXAMINATION) listed before E.

Hence after fixing one A as first word we can rearrange the remaining 10 letters of which 2 are I, 2 are N and rest are all different.

Therefore, the required number of ways = $\frac{10!}{2! 2!}$

= 10 * 9 * 8 * 7 * 6 * 5 * 2 * 3

= 9,07,200

<p><strong>34.</strong> In the eleven-letter word EXAMINATION there are 2A's, 2I's and 2N's and the rest are all different.</p><p>Since in a dictionary the words are listed according to the English alphabet we can only find words starting with A (as B, C, D are not a part of the letters forming the word EXAMINATION) listed before E.</p><p>Hence after fixing one A as first word we can rearrange the remaining 10 letters of which 2 are I, 2 are N and rest are all different.</p><p>Therefore, the required number of ways =&nbsp;<span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>1</mn><mn>0</mn><mo>!</mo></mrow><mrow><mn>2</mn><mo>!</mo><mo>&nbsp;</mo><mn>2</mn><mo>!</mo></mrow></mfrac></mrow></math></span></p><div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1747821231phpUdRXCd.jpeg" alt="" width="299" height="62" loading="lazy"></picture></div></div><p>= 10 * 9 * 8 * 7 * 6 * 5 * 2 * 3</p><p>= 9,07,200</p>

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